A progressive collection of 20 practice problems on exponential equations, designed to help you not only find the answer, but above all to recognize which method to apply. In each exercise we will analyze the structure of the equation, identify the most convenient transformation, and carry out every step without skipping any intermediate work.
Recall the fundamental principle: if \(a>0\) and \(a\ne1\), then:
\[ a^u=a^v \quad \Longleftrightarrow \quad u=v \]
This property means that when two powers share the same positive base (other than \(1\)), the equality of the powers is equivalent to the equality of the exponents. For this reason, in many exercises the first objective will be to rewrite both sides as powers with the same base.
Exercise 1 — level ★☆☆☆☆
Solve the equation:
\[ 2^x=16 \]
Answer
\[ S=\{4\} \]
Solution
This is an exponential equation because the unknown \(x\) appears in the exponent:
\[ 2^x=16 \]
The left-hand side is already a power with base \(2\). In order to compare the exponents, we need to write the right-hand side as a power of \(2\) as well.
We observe that:
\[ 16=2^4 \]
since:
\[ 2^4=2\cdot2\cdot2\cdot2=16 \]
Replacing \(16\) with \(2^4\), the equation becomes:
\[ 2^x=2^4 \]
Both sides are now powers with the same base \(2\). Since:
\[ 2>0 \quad \text{and} \quad 2\ne1 \]
we can apply the injectivity of the exponential function and equate the exponents:
\[ x=4 \]
Therefore:
\[ S=\{4\} \]
Exercise 2 — level ★☆☆☆☆
Solve the equation:
\[ 3^x=81 \]
Answer
\[ S=\{4\} \]
Solution
This is also an exponential equation, since the unknown \(x\) appears in the exponent.
The left-hand side is a power with base \(3\):
\[ 3^x \]
To use the same-base method, we need to rewrite the right-hand side as a power of \(3\) as well.
Let us compute successive powers of \(3\):
\[ 3^1=3,\qquad 3^2=9,\qquad 3^3=27,\qquad 3^4=81 \]
Hence:
\[ 81=3^4 \]
The equation can therefore be rewritten as:
\[ 3^x=3^4 \]
The bases are now equal. We are not cancelling the \(3\) mechanically: we are using the fact that the exponential function with base \(3\) is injective.
Therefore the exponents must be equal:
\[ x=4 \]
The solution is:
\[ S=\{4\} \]
Exercise 3 — level ★☆☆☆☆
Solve the equation:
\[ 5^{x-1}=25 \]
Answer
\[ S=\{3\} \]
Solution
The left-hand side is a power with base \(5\), but the exponent is not simply \(x\): it is \(x-1\).
This does not change the method. We still need to write the right-hand side as a power of the same base.
Since:
\[ 25=5^2 \]
we can rewrite the equation as:
\[ 5^{x-1}=5^2 \]
Both sides are now powers with the same base \(5\). Since:
\[ 5>0 \quad \text{and} \quad 5\ne1 \]
we can equate the exponents:
\[ x-1=2 \]
This is no longer an exponential equation, but a simple linear equation. Adding \(1\) to both sides:
\[ x=2+1 \]
hence:
\[ x=3 \]
Therefore:
\[ S=\{3\} \]
Exercise 4 — level ★★☆☆☆
Solve the equation:
\[ 2^{2x+1}=32 \]
Answer
\[ S=\{2\} \]
Solution
The equation contains a power with base \(2\):
\[ 2^{2x+1} \]
The right-hand side is the number \(32\). Before we can compare the exponents, we need to write \(32\) as a power of \(2\).
Since:
\[ 32=2^5 \]
the equation becomes:
\[ 2^{2x+1}=2^5 \]
Both powers share the same positive base, which is not equal to \(1\). We can therefore equate the exponents:
\[ 2x+1=5 \]
We now solve the resulting linear equation. Subtracting \(1\) from both sides:
\[ 2x=5-1 \]
hence:
\[ 2x=4 \]
Dividing both sides by \(2\):
\[ x=2 \]
Therefore:
\[ S=\{2\} \]
Exercise 5 — level ★★☆☆☆
Solve the equation:
\[ 4^x=64 \]
Answer
\[ S=\{3\} \]
Solution
The left-hand side is a power with base \(4\):
\[ 4^x \]
To apply the same-base method, we need to write the right-hand side as a power of \(4\) as well.
We observe that:
\[ 64=4^3 \]
since:
\[ 4^3=4\cdot4\cdot4=64 \]
The equation therefore becomes:
\[ 4^x=4^3 \]
Both sides are now powers with the same base \(4\). Since \(4>0\) and \(4\ne1\), we can equate the exponents:
\[ x=3 \]
Therefore:
\[ S=\{3\} \]
Exercise 6 — level ★★☆☆☆
Solve the equation:
\[ 9^x=3^{x+2} \]
Answer
\[ S=\{2\} \]
Solution
In this equation the bases do not match: the left-hand side has base \(9\), while the right-hand side has base \(3\).
However, \(9\) can be written as a power of \(3\):
\[ 9=3^2 \]
We therefore rewrite the left-hand side:
\[ 9^x=(3^2)^x \]
Applying the power of a power rule:
\[ (a^m)^n=a^{mn} \]
we obtain:
\[ (3^2)^x=3^{2x} \]
The original equation therefore becomes:
\[ 3^{2x}=3^{x+2} \]
Both sides are now powers with the same positive base, not equal to \(1\). We can equate the exponents:
\[ 2x=x+2 \]
Subtracting \(x\) from both sides:
\[ 2x-x=2 \]
hence:
\[ x=2 \]
Therefore:
\[ S=\{2\} \]
Exercise 7 — level ★★☆☆☆
Solve the equation:
\[ 8^x=4^{x+1} \]
Answer
\[ S=\{2\} \]
Solution
This equation involves two different bases:
\[ 8 \quad \text{and} \quad 4 \]
Before comparing the exponents, we need to find a common base.
We observe that both \(8\) and \(4\) are powers of \(2\):
\[ 8=2^3 \]
and:
\[ 4=2^2 \]
We now rewrite both sides accordingly.
For the left-hand side:
\[ 8^x=(2^3)^x \]
Applying the power of a power rule:
\[ (2^3)^x=2^{3x} \]
For the right-hand side:
\[ 4^{x+1}=(2^2)^{x+1} \]
Applying the same rule again:
\[ (2^2)^{x+1}=2^{2(x+1)} \]
The equation becomes:
\[ 2^{3x}=2^{2(x+1)} \]
The bases now match, so we equate the exponents:
\[ 3x=2(x+1) \]
Expanding the right-hand side:
\[ 3x=2x+2 \]
Subtracting \(2x\) from both sides:
\[ 3x-2x=2 \]
hence:
\[ x=2 \]
Therefore:
\[ S=\{2\} \]
Exercise 8 — level ★★☆☆☆
Solve the equation:
\[ 2^{x+1}\cdot2^{x-2}=8 \]
Answer
\[ S=\{2\} \]
Solution
The left-hand side is a product of two powers with the same base:
\[ 2^{x+1}\cdot2^{x-2} \]
When multiplying powers that share the same base, we keep the base and add the exponents:
\[ a^m\cdot a^n=a^{m+n} \]
Applying this property:
\[ 2^{x+1}\cdot2^{x-2}=2^{(x+1)+(x-2)} \]
We now simplify the exponent:
\[ (x+1)+(x-2)=x+1+x-2=2x-1 \]
The left-hand side therefore becomes:
\[ 2^{2x-1} \]
The equation now reads:
\[ 2^{2x-1}=8 \]
We write \(8\) as a power of \(2\):
\[ 8=2^3 \]
giving:
\[ 2^{2x-1}=2^3 \]
Since the bases match, we equate the exponents:
\[ 2x-1=3 \]
Adding \(1\) to both sides:
\[ 2x=4 \]
Dividing both sides by \(2\):
\[ x=2 \]
The solution is:
\[ S=\{2\} \]
Exercise 9 — level ★★★☆☆
Solve the equation:
\[ \frac{3^{x+2}}{3^{x-1}}=27 \]
Answer
\[ S=\mathbb{R} \]
Solution
The left-hand side is a quotient of powers with the same base:
\[ \frac{3^{x+2}}{3^{x-1}} \]
When dividing powers with the same base, we keep the base and subtract the exponents:
\[ \frac{a^m}{a^n}=a^{m-n} \qquad (a\ne0) \]
Applying this property:
\[ \frac{3^{x+2}}{3^{x-1}} = 3^{(x+2)-(x-1)} \]
We simplify the exponent carefully:
\[ (x+2)-(x-1)=x+2-x+1 \]
hence:
\[ (x+2)-(x-1)=3 \]
The left-hand side therefore reduces to:
\[ 3^3 \]
Since:
\[ 3^3=27 \]
the original equation reduces to:
\[ 27=27 \]
This identity holds for every real number, and imposes no condition on the unknown \(x\).
Every real number is therefore a solution.
Therefore:
\[ S=\mathbb{R} \]
Exercise 10 — level ★★★☆☆
Solve the equation:
\[ 5^{2x}=125\cdot5^x \]
Answer
\[ S=\{3\} \]
Solution
The right-hand side contains the number \(125\) multiplied by a power of \(5\). To work with a single base, we rewrite \(125\) as a power of \(5\).
Since:
\[ 125=5^3 \]
we obtain:
\[ 125\cdot5^x=5^3\cdot5^x \]
Now we use the product rule for powers with the same base:
\[ a^m\cdot a^n=a^{m+n} \]
Therefore:
\[ 5^3\cdot5^x=5^{3+x} \]
that is:
\[ 5^3\cdot5^x=5^{x+3} \]
The original equation becomes:
\[ 5^{2x}=5^{x+3} \]
The bases now match, so we equate the exponents:
\[ 2x=x+3 \]
Subtracting \(x\) from both sides:
\[ 2x-x=3 \]
hence:
\[ x=3 \]
Therefore:
\[ S=\{3\} \]
Exercise 11 — level ★★★☆☆
Solve the equation:
\[ 2^{2x}-5\cdot2^x+4=0 \]
Answer
\[ S=\{0,2\} \]
Solution
In this equation we cannot directly equate bases, because the unknown appears in more than one term:
\[ 2^{2x}-5\cdot2^x+4=0 \]
However, we notice an important structure:
\[ 2^{2x}=(2^x)^2 \]
This means the equation can be interpreted as a quadratic in the quantity \(2^x\).
We therefore introduce the substitution:
\[ t=2^x \]
Since a power with a positive base is always positive, we require:
\[ t>0 \]
Substituting into the equation gives:
\[ t^2-5t+4=0 \]
This is now an ordinary quadratic equation, not an exponential one.
We look for two numbers whose product is \(4\) and whose sum is \(-5\). These numbers are \(-1\) and \(-4\).
We can therefore factor the trinomial:
\[ t^2-5t+4=(t-1)(t-4) \]
The equation becomes:
\[ (t-1)(t-4)=0 \]
A product is zero when at least one factor is zero. We obtain:
\[ t-1=0 \]
or:
\[ t-4=0 \]
giving:
\[ t=1 \]
or:
\[ t=4 \]
Both values are positive, so both satisfy the condition \(t>0\).
We now return to the original variable.
If:
\[ t=1 \]
then:
\[ 2^x=1 \]
Recalling that:
\[ 1=2^0 \]
we get:
\[ 2^x=2^0 \]
hence:
\[ x=0 \]
If instead:
\[ t=4 \]
then:
\[ 2^x=4 \]
Since:
\[ 4=2^2 \]
we obtain:
\[ 2^x=2^2 \]
hence:
\[ x=2 \]
The solutions are:
\[ S=\{0,2\} \]
Exercise 12 — level ★★★☆☆
Solve the equation:
\[ 3^{2x}-10\cdot3^x+9=0 \]
Answer
\[ S=\{0,2\} \]
Solution
This equation also has the structure of a quadratic trinomial.
Indeed:
\[ 3^{2x}=(3^x)^2 \]
We introduce the substitution:
\[ t=3^x \]
Since a positive power is always positive:
\[ t>0 \]
Substituting gives:
\[ t^2-10t+9=0 \]
We look for two numbers whose product is \(9\) and whose sum is \(-10\). These numbers are \(-1\) and \(-9\).
We can therefore factor:
\[ t^2-10t+9=(t-1)(t-9) \]
The equation becomes:
\[ (t-1)(t-9)=0 \]
A product is zero when at least one factor is zero. We obtain:
\[ t=1 \]
or:
\[ t=9 \]
Both solutions satisfy the condition \(t>0\).
We now return to the original variable.
If:
\[ t=1 \]
then:
\[ 3^x=1 \]
Since:
\[ 1=3^0 \]
we obtain:
\[ 3^x=3^0 \]
hence:
\[ x=0 \]
If instead:
\[ t=9 \]
then:
\[ 3^x=9 \]
Since:
\[ 9=3^2 \]
we obtain:
\[ 3^x=3^2 \]
hence:
\[ x=2 \]
The solutions are:
\[ S=\{0,2\} \]
Exercise 13 — level ★★★☆☆
Solve the equation:
\[ 4^x-6\cdot2^x+8=0 \]
Answer
\[ S=\{1,2\} \]
Solution
This equation contains both \(4^x\) and \(2^x\). To make a substitution possible, we first need to express everything in terms of the same base.
We observe that:
\[ 4=2^2 \]
therefore:
\[ 4^x=(2^2)^x \]
Applying the power of a power rule:
\[ (2^2)^x=2^{2x} \]
Furthermore:
\[ 2^{2x}=(2^x)^2 \]
The original equation therefore becomes:
\[ (2^x)^2-6\cdot2^x+8=0 \]
We now introduce the substitution:
\[ t=2^x \]
with the condition:
\[ t>0 \]
This gives:
\[ t^2-6t+8=0 \]
We look for two numbers whose product is \(8\) and whose sum is \(-6\). These numbers are \(-2\) and \(-4\).
We can therefore factor:
\[ t^2-6t+8=(t-2)(t-4) \]
The equation becomes:
\[ (t-2)(t-4)=0 \]
giving:
\[ t=2 \]
or:
\[ t=4 \]
Both solutions are positive, so both are admissible.
We now return to the original variable.
If:
\[ t=2 \]
then:
\[ 2^x=2 \]
that is:
\[ 2^x=2^1 \]
hence:
\[ x=1 \]
If instead:
\[ t=4 \]
then:
\[ 2^x=4 \]
Since:
\[ 4=2^2 \]
we obtain:
\[ 2^x=2^2 \]
hence:
\[ x=2 \]
The solutions are:
\[ S=\{1,2\} \]
Exercise 14 — level ★★★★☆
Solve the equation:
\[ 2^{x+1}+2^x=24 \]
Answer
\[ S=\{3\} \]
Solution
This equation contains two powers with the same base \(2\), but with different exponents:
\[ 2^{x+1} \quad \text{and} \quad 2^x \]
The idea is to rewrite both powers in terms of the same quantity, namely \(2^x\).
We observe that:
\[ 2^{x+1}=2^x\cdot2^1 \]
Since:
\[ 2^1=2 \]
we get:
\[ 2^{x+1}=2\cdot2^x \]
Substituting this into the original equation:
\[ 2\cdot2^x+2^x=24 \]
Both terms on the left-hand side share the common factor \(2^x\). Factoring it out:
\[ 2^x(2+1)=24 \]
Computing the sum in parentheses:
\[ 2+1=3 \]
The equation becomes:
\[ 3\cdot2^x=24 \]
Dividing both sides by \(3\):
\[ 2^x=8 \]
Writing \(8\) as a power of \(2\):
\[ 8=2^3 \]
we obtain:
\[ 2^x=2^3 \]
Since the bases match, we equate the exponents:
\[ x=3 \]
The solution is:
\[ S=\{3\} \]
Exercise 15 — level ★★★★☆
Solve the equation:
\[ 3^{x+2}-3^x=72 \]
Answer
\[ S=\{2\} \]
Solution
This equation contains two powers with the same base \(3\), but with different exponents:
\[ 3^{x+2} \quad \text{and} \quad 3^x \]
The idea is to rewrite \(3^{x+2}\) so as to factor out \(3^x\).
Using the property:
\[ a^{m+n}=a^m\cdot a^n \]
we can write:
\[ 3^{x+2}=3^x\cdot3^2 \]
Since:
\[ 3^2=9 \]
it follows that:
\[ 3^{x+2}=9\cdot3^x \]
Substituting into the original equation:
\[ 9\cdot3^x-3^x=72 \]
Both terms on the left-hand side share the common factor \(3^x\). Factoring it out:
\[ 3^x(9-1)=72 \]
Computing:
\[ 9-1=8 \]
Therefore:
\[ 8\cdot3^x=72 \]
Dividing both sides by \(8\):
\[ 3^x=9 \]
Writing \(9\) as a power of \(3\):
\[ 9=3^2 \]
we obtain:
\[ 3^x=3^2 \]
Since the bases match, we equate the exponents:
\[ x=2 \]
Therefore:
\[ S=\{2\} \]
Exercise 16 — level ★★★★☆
Solve the equation:
\[ 2^{x+2}-2^{x+1}=8 \]
Answer
\[ S=\{2\} \]
Solution
This equation contains two powers with base \(2\):
\[ 2^{x+2} \quad \text{and} \quad 2^{x+1} \]
To simplify the expression, it is convenient to rewrite both in terms of \(2^x\).
For the first power:
\[ 2^{x+2}=2^x\cdot2^2 \]
Since:
\[ 2^2=4 \]
we obtain:
\[ 2^{x+2}=4\cdot2^x \]
For the second power:
\[ 2^{x+1}=2^x\cdot2^1 \]
hence:
\[ 2^{x+1}=2\cdot2^x \]
Substituting these expressions into the original equation:
\[ 4\cdot2^x-2\cdot2^x=8 \]
Factoring out the common factor \(2^x\):
\[ 2^x(4-2)=8 \]
Computing:
\[ 4-2=2 \]
Therefore:
\[ 2\cdot2^x=8 \]
Dividing both sides by \(2\):
\[ 2^x=4 \]
Writing \(4\) as a power of \(2\):
\[ 4=2^2 \]
we get:
\[ 2^x=2^2 \]
Equating the exponents:
\[ x=2 \]
The solution is:
\[ S=\{2\} \]
Exercise 17 — level ★★★★☆
Solve the equation:
\[ 2^x+2^{-x}=\frac{5}{2} \]
Answer
\[ S=\{-1,1\} \]
Solution
This equation contains two related powers:
\[ 2^x \quad \text{and} \quad 2^{-x} \]
The presence of the negative exponent suggests using the property:
\[ a^{-n}=\frac{1}{a^n} \]
Applying it:
\[ 2^{-x}=\frac{1}{2^x} \]
The equation becomes:
\[ 2^x+\frac{1}{2^x}=\frac{5}{2} \]
The quantity \(2^x\) now appears throughout. We therefore introduce the substitution:
\[ t=2^x \]
Since a power with a positive base is always positive:
\[ t>0 \]
Also:
\[ \frac{1}{2^x}=\frac{1}{t} \]
The equation transforms into:
\[ t+\frac{1}{t}=\frac{5}{2} \]
To clear the denominator, we multiply both sides by \(2t\).
This is valid because \(t>0\), so:
\[ 2t\ne0 \]
We obtain:
\[ 2t\left(t+\frac{1}{t}\right)=2t\cdot\frac{5}{2} \]
Expanding the left-hand side:
\[ 2t\cdot t+2t\cdot\frac{1}{t}=5t \]
that is:
\[ 2t^2+2=5t \]
Bringing all terms to the left-hand side:
\[ 2t^2-5t+2=0 \]
Factoring the trinomial:
\[ 2t^2-5t+2=(2t-1)(t-2) \]
The equation becomes:
\[ (2t-1)(t-2)=0 \]
By the zero product property:
\[ 2t-1=0 \]
or:
\[ t-2=0 \]
In the first case:
\[ 2t=1 \]
hence:
\[ t=\frac{1}{2} \]
In the second case:
\[ t=2 \]
Both values satisfy the condition \(t>0\).
We now return to the original variable.
If:
\[ t=\frac{1}{2} \]
then:
\[ 2^x=\frac{1}{2} \]
Since:
\[ \frac{1}{2}=2^{-1} \]
we obtain:
\[ 2^x=2^{-1} \]
hence:
\[ x=-1 \]
If instead:
\[ t=2 \]
then:
\[ 2^x=2 \]
that is:
\[ 2^x=2^1 \]
hence:
\[ x=1 \]
The solutions are:
\[ S=\{-1,1\} \]
Exercise 18 — level ★★★★☆
Solve the equation:
\[ 3^x=7 \]
Answer
\[ S=\{\log_3 7\} \]
Solution
In this equation the unknown \(x\) appears in the exponent:
\[ 3^x=7 \]
The left-hand side is a power with base \(3\). To use the same-base method, we would need to write \(7\) as a power of \(3\) as well.
However, \(7\) is not an integer power of \(3\). Indeed:
\[ 3^1=3 \]
while:
\[ 3^2=9 \]
The number \(7\) lies between \(3\) and \(9\), but does not coincide with any integer power of \(3\).
This does not mean the equation has no solution. It simply means the solution cannot be obtained as a simple integer exponent.
To find the exponent to which \(3\) must be raised in order to get \(7\), we use the logarithm base \(3\).
By definition:
\[ \log_3 7 \]
is precisely the exponent to which \(3\) must be raised to obtain \(7\).
Therefore:
\[ 3^x=7 \quad \Longleftrightarrow \quad x=\log_3 7 \]
The solution is:
\[ S=\{\log_3 7\} \]
Exercise 19 — level ★★★★★
Solve the equation:
\[ 2^{3x-1}=5 \]
Answer
\[ S=\left\{\frac{1+\log_2 5}{3}\right\} \]
Solution
This is an exponential equation because the unknown appears in the exponent:
\[ 2^{3x-1}=5 \]
The left-hand side is a power with base \(2\). If the right-hand side were a power of \(2\), we could equate the exponents directly.
However, \(5\) is not an integer power of \(2\). Indeed:
\[ 2^2=4 \]
while:
\[ 2^3=8 \]
The number \(5\) lies between \(4\) and \(8\), so a solution exists, but it is not an integer.
To isolate the exponent \(3x-1\), we apply the logarithm base \(2\) to both sides:
\[ \log_2\left(2^{3x-1}\right)=\log_2 5 \]
The logarithm base \(2\) and the exponential base \(2\) are inverse operations. Therefore:
\[ \log_2\left(2^{3x-1}\right)=3x-1 \]
The equation becomes:
\[ 3x-1=\log_2 5 \]
This is now a simple linear equation in \(x\).
Adding \(1\) to both sides:
\[ 3x=1+\log_2 5 \]
Dividing both sides by \(3\):
\[ x=\frac{1+\log_2 5}{3} \]
The solution is:
\[ S=\left\{\frac{1+\log_2 5}{3}\right\} \]
Exercise 20 — level ★★★★★
Solve the equation:
\[ 4^x+2^x-6=0 \]
Answer
\[ S=\{1\} \]
Solution
This equation contains two different powers:
\[ 4^x \quad \text{and} \quad 2^x \]
The presence of both \(4^x\) and \(2^x\) suggests rewriting everything in terms of the same quantity.
Since:
\[ 4=2^2 \]
we can transform \(4^x\) as follows:
\[ 4^x=(2^2)^x \]
Applying the power of a power rule:
\[ (2^2)^x=2^{2x} \]
Furthermore:
\[ 2^{2x}=(2^x)^2 \]
Therefore:
\[ 4^x=(2^x)^2 \]
The original equation:
\[ 4^x+2^x-6=0 \]
becomes:
\[ (2^x)^2+2^x-6=0 \]
The structure now resembles a quadratic equation. We therefore introduce the substitution:
\[ t=2^x \]
Since a power with a positive base is always positive:
\[ t>0 \]
Substituting gives:
\[ t^2+t-6=0 \]
We look for two numbers whose product is \(-6\) and whose sum is \(1\). These numbers are \(3\) and \(-2\).
We can therefore factor:
\[ t^2+t-6=(t+3)(t-2) \]
The equation becomes:
\[ (t+3)(t-2)=0 \]
By the zero product property, at least one factor must be zero. We obtain:
\[ t+3=0 \]
or:
\[ t-2=0 \]
In the first case:
\[ t=-3 \]
In the second case:
\[ t=2 \]
Recall, however, the condition from our substitution:
\[ t>0 \]
The value:
\[ t=-3 \]
must be rejected, since no real number \(x\) satisfies:
\[ 2^x=-3 \]
Only the value:
\[ t=2 \]
remains. Returning to the original variable, since:
\[ t=2^x \]
the condition \(t=2\) gives:
\[ 2^x=2 \]
that is:
\[ 2^x=2^1 \]
Since the bases match, we equate the exponents:
\[ x=1 \]
The solution is:
\[ S=\{1\} \]