A graded collection of 20 practice problems on exponential inequalities, designed to build a solid understanding of how to apply the monotonicity of exponential functions correctly — recognizing when the direction of an inequality is preserved and when it must be reversed.
In each exercise we carefully apply the laws of exponents, reduce both sides to a common base, and, where necessary, employ the substitution method.
Exercise 1 — level ★☆☆☆☆
Solve:
\[ 2^x>8 \]
Answer
\[ S=(3,+\infty) \]
Solution
Write \(8\) as a power of \(2\):
\[ 8=2^3 \]
The inequality becomes:
\[ 2^x>2^3 \]
Since \(2>1\), the exponential function \(2^x\) is strictly increasing. We may therefore compare exponents while preserving the direction of the inequality:
\[ x>3 \]
Therefore:
\[ S=(3,+\infty) \]
Exercise 2 — level ★☆☆☆☆
Solve:
\[ 3^{x-1}\le 27 \]
Answer
\[ S=(-\infty,4] \]
Solution
Write \(27\) as a power of \(3\):
\[ 27=3^3 \]
We obtain:
\[ 3^{x-1}\le 3^3 \]
Since \(3>1\), the exponential function is increasing. The direction of the inequality is preserved:
\[ x-1\le 3 \]
Hence:
\[ x\le 4 \]
Therefore:
\[ S=(-\infty,4] \]
Exercise 3 — level ★☆☆☆☆
Solve:
\[ \left(\frac12\right)^x>\frac1{16} \]
Answer
\[ S=(-\infty,4) \]
Solution
Write the right-hand side as a power of \(\frac12\):
\[ \frac1{16}=\left(\frac12\right)^4 \]
The inequality becomes:
\[ \left(\frac12\right)^x>\left(\frac12\right)^4 \]
Since:
\[ 0<\frac12<1 \]
the exponential function is strictly decreasing. Consequently, when we compare exponents, the direction of the inequality reverses:
\[ x<4 \]
Therefore:
\[ S=(-\infty,4) \]
Exercise 4 — level ★★☆☆☆
Solve:
\[ 5^{2x+1}\ge 5^{x-3} \]
Answer
\[ S=[-4,+\infty) \]
Solution
Both sides already share the base \(5\). Since \(5>1\), the exponential function is increasing.
We may therefore compare exponents while preserving the direction of the inequality:
\[ 2x+1\ge x-3 \]
Subtracting \(x\) from both sides:
\[ x+1\ge -3 \]
Hence:
\[ x\ge -4 \]
Therefore:
\[ S=[-4,+\infty) \]
Exercise 5 — level ★★☆☆☆
Solve:
\[ \left(\frac13\right)^{x+2}\le \left(\frac13\right)^{2x-1} \]
Answer
\[ S=(-\infty,3] \]
Solution
The base is \(\frac13\), so:
\[ 0<\frac13<1 \]
The exponential function is therefore decreasing. Accordingly, when we pass from the exponentials to the exponents, the direction of the inequality reverses.
From:
\[ \left(\frac13\right)^{x+2}\le \left(\frac13\right)^{2x-1} \]
we obtain:
\[ x+2\ge 2x-1 \]
Hence:
\[ 3\ge x \]
that is:
\[ x\le 3 \]
Therefore:
\[ S=(-\infty,3] \]
Exercise 6 — level ★★☆☆☆
Solve:
\[ 4^x>2^{3x} \]
Answer
\[ S=(-\infty,0) \]
Solution
Write \(4\) as a power of \(2\):
\[ 4=2^2 \]
Then:
\[ 4^x=(2^2)^x=2^{2x} \]
The inequality becomes:
\[ 2^{2x}>2^{3x} \]
Since \(2>1\), the exponential function \(2^x\) is strictly increasing. We may therefore compare exponents while preserving the direction of the inequality:
\[ 2x>3x \]
Subtracting \(3x\) from both sides:
\[ -x>0 \]
Multiplying both sides by \(-1\) reverses the direction of the inequality:
\[ x<0 \]
Therefore:
\[ S=(-\infty,0) \]
Exercise 7 — level ★★☆☆☆
Solve:
\[ 9^x\le 3^{x+4} \]
Answer
\[ S=(-\infty,4] \]
Solution
Write \(9\) as a power of \(3\):
\[ 9=3^2 \]
Then:
\[ 9^x=(3^2)^x=3^{2x} \]
The inequality becomes:
\[ 3^{2x}\le 3^{x+4} \]
Since \(3>1\), we compare exponents while preserving the direction of the inequality:
\[ 2x\le x+4 \]
Hence:
\[ x\le 4 \]
Therefore:
\[ S=(-\infty,4] \]
Exercise 8 — level ★★☆☆☆
Solve:
\[ 2^{x+2}-2^x>12 \]
Answer
\[ S=(2,+\infty) \]
Solution
Rewrite the first term:
\[ 2^{x+2}=2^x\cdot 2^2=4\cdot 2^x \]
The inequality becomes:
\[ 4\cdot 2^x-2^x>12 \]
Factor out \(2^x\):
\[ 2^x(4-1)>12 \]
that is:
\[ 3\cdot 2^x>12 \]
Dividing by \(3\), which is positive:
\[ 2^x>4 \]
Since \(4=2^2\), we obtain:
\[ 2^x>2^2 \]
Since \(2>1\), the exponential function is increasing:
\[ x>2 \]
Therefore:
\[ S=(2,+\infty) \]
Exercise 9 — level ★★☆☆☆
Solve:
\[ 3^{x+1}+3^x\le 36 \]
Answer
\[ S=(-\infty,2] \]
Solution
Rewrite:
\[ 3^{x+1}=3\cdot 3^x \]
Therefore:
\[ 3^{x+1}+3^x=3\cdot 3^x+3^x=4\cdot 3^x \]
The inequality becomes:
\[ 4\cdot 3^x\le 36 \]
Dividing by \(4\), which is positive:
\[ 3^x\le 9 \]
Since \(9=3^2\), we obtain:
\[ 3^x\le 3^2 \]
Since \(3>1\), we compare exponents:
\[ x\le 2 \]
Therefore:
\[ S=(-\infty,2] \]
Exercise 10 — level ★★★☆☆
Solve:
\[ 2^{2x}-5\cdot 2^x+4\le 0 \]
Answer
\[ S=[0,2] \]
Solution
Let:
\[ t=2^x \]
Since \(2^x>0\), we have:
\[ t>0 \]
Moreover:
\[ 2^{2x}=(2^x)^2=t^2 \]
The inequality becomes:
\[ t^2-5t+4\le 0 \]
Factoring:
\[ t^2-5t+4=(t-1)(t-4) \]
So:
\[ (t-1)(t-4)\le 0 \]
The product is less than or equal to zero between the two roots:
\[ 1\le t\le 4 \]
Back-substituting \(t=2^x\):
\[ 1\le 2^x\le 4 \]
Writing:
\[ 1=2^0,\qquad 4=2^2 \]
and using the fact that \(2>1\), we obtain:
\[ 0\le x\le 2 \]
Therefore:
\[ S=[0,2] \]
Exercise 11 — level ★★★☆☆
Solve:
\[ 3^{2x}-4\cdot 3^x+3>0 \]
Answer
\[ S=(-\infty,0)\cup(1,+\infty) \]
Solution
Let:
\[ t=3^x \]
Since \(3^x>0\), we have:
\[ t>0 \]
Moreover:
\[ 3^{2x}=(3^x)^2=t^2 \]
The inequality becomes:
\[ t^2-4t+3>0 \]
Factoring:
\[ t^2-4t+3=(t-1)(t-3) \]
So:
\[ (t-1)(t-3)>0 \]
The product is positive outside the two roots:
\[ t<1 \quad \text{or} \quad t>3 \]
Taking into account that \(t>0\), the first condition becomes:
\[ 0
Back-substituting \(t=3^x\):
\[ 3^x<1 \quad \text{or} \quad 3^x>3 \]
Writing:
\[ 1=3^0,\qquad 3=3^1 \]
and using the fact that \(3>1\), we obtain:
\[ x<0 \quad \text{or} \quad x>1 \]
Therefore:
\[ S=(-\infty,0)\cup(1,+\infty) \]
Exercise 12 — level ★★★☆☆
Solve:
\[ 4^x-6\cdot 2^x+8\ge 0 \]
Answer
\[ S=(-\infty,1]\cup[2,+\infty) \]
Solution
Observe that:
\[ 4^x=(2^2)^x=2^{2x}=(2^x)^2 \]
Let:
\[ t=2^x \]
with:
\[ t>0 \]
The inequality becomes:
\[ t^2-6t+8\ge 0 \]
Factoring:
\[ t^2-6t+8=(t-2)(t-4) \]
So:
\[ (t-2)(t-4)\ge 0 \]
The product is greater than or equal to zero outside the two roots:
\[ t\le 2 \quad \text{or} \quad t\ge 4 \]
Back-substituting \(t=2^x\):
\[ 2^x\le 2 \quad \text{or} \quad 2^x\ge 4 \]
Since:
\[ 2=2^1,\qquad 4=2^2 \]
and \(2>1\), we obtain:
\[ x\le 1 \quad \text{or} \quad x\ge 2 \]
Therefore:
\[ S=(-\infty,1]\cup[2,+\infty) \]
Exercise 13 — level ★★★☆☆
Solve:
\[ \frac{2^x-4}{2^x+1}\ge 0 \]
Answer
\[ S=[2,+\infty) \]
Solution
Let:
\[ t=2^x \]
Since \(2^x>0\), we have:
\[ t>0 \]
The inequality becomes:
\[ \frac{t-4}{t+1}\ge 0 \]
Since \(t>0\), the denominator is always positive:
\[ t+1>0 \]
Hence the sign of the fraction is determined entirely by the numerator:
\[ t-4\ge 0 \]
that is:
\[ t\ge 4 \]
Back-substituting \(t=2^x\):
\[ 2^x\ge 4 \]
Since \(4=2^2\) and \(2>1\), we obtain:
\[ x\ge 2 \]
Therefore:
\[ S=[2,+\infty) \]
Exercise 14 — level ★★★★☆
Solve:
\[ \frac{3^x-9}{3^x-1}<0 \]
Answer
\[ S=(0,2) \]
Solution
Let:
\[ t=3^x \]
Since \(3^x>0\), we have:
\[ t>0 \]
The inequality becomes:
\[ \frac{t-9}{t-1}<0 \]
The critical values are:
\[ t=1,\qquad t=9 \]
The value \(t=1\) makes the denominator zero and is therefore excluded. The value \(t=9\) makes the numerator zero.
For \(t>0\), we analyze the sign on each subinterval:
\[ (0,1),\qquad (1,9),\qquad (9,+\infty) \]
The sign chart is:
\[ \begin{array}{c|ccc} t & (0,1) & (1,9) & (9,+\infty)\\ \hline t-9 & - & - & +\\ t-1 & - & + & +\\ \hline \dfrac{t-9}{t-1} & + & - & + \end{array} \]
We need the fraction to be negative, so:
\[ 1
Back-substituting \(t=3^x\):
\[ 1<3^x<9 \]
Writing:
\[ 1=3^0,\qquad 9=3^2 \]
So:
\[ 3^0<3^x<3^2 \]
Since \(3>1\), we obtain:
\[ 0
Therefore:
\[ S=(0,2) \]
Exercise 15 — level ★★★★☆
Solve:
\[ 2^{x+1}+2^{1-x}\le 5 \]
Answer
\[ S=[-1,1] \]
Solution
Let:
\[ t=2^x \]
Since \(2^x>0\), we have:
\[ t>0 \]
Rewrite each term:
\[ 2^{x+1}=2\cdot 2^x=2t \]
and:
\[ 2^{1-x}=2\cdot 2^{-x}=\frac{2}{2^x}=\frac{2}{t} \]
The inequality becomes:
\[ 2t+\frac{2}{t}\le 5 \]
Since \(t>0\), we may multiply through by \(t\) without reversing the inequality:
\[ 2t^2+2\le 5t \]
Bringing all terms to the left-hand side:
\[ 2t^2-5t+2\le 0 \]
Factoring:
\[ 2t^2-5t+2=(2t-1)(t-2) \]
So:
\[ (2t-1)(t-2)\le 0 \]
The product is less than or equal to zero between the two roots:
\[ \frac12\le t\le 2 \]
Back-substituting \(t=2^x\):
\[ \frac12\le 2^x\le 2 \]
Writing:
\[ \frac12=2^{-1},\qquad 2=2^1 \]
and using the fact that \(2>1\), we obtain:
\[ -1\le x\le 1 \]
Therefore:
\[ S=[-1,1] \]
Exercise 16 — level ★★★★☆
Solve:
\[ 9^x-10\cdot 3^x+9\ge 0 \]
Answer
\[ S=(-\infty,0]\cup[2,+\infty) \]
Solution
Observe that:
\[ 9^x=(3^2)^x=3^{2x}=(3^x)^2 \]
Let:
\[ t=3^x \]
with:
\[ t>0 \]
The inequality becomes:
\[ t^2-10t+9\ge 0 \]
Factoring:
\[ t^2-10t+9=(t-1)(t-9) \]
So:
\[ (t-1)(t-9)\ge 0 \]
The product is greater than or equal to zero outside the two roots:
\[ t\le 1 \quad \text{or} \quad t\ge 9 \]
Back-substituting \(t=3^x\):
\[ 3^x\le 1 \quad \text{or} \quad 3^x\ge 9 \]
Since:
\[ 1=3^0,\qquad 9=3^2 \]
and \(3>1\), we obtain:
\[ x\le 0 \quad \text{or} \quad x\ge 2 \]
Therefore:
\[ S=(-\infty,0]\cup[2,+\infty) \]
Exercise 17 — level ★★★★☆
Solve:
\[ \left(\frac14\right)^x-5\left(\frac12\right)^x+4\le 0 \]
Answer
\[ S=[-2,0] \]
Solution
We rewrite everything in terms of \(\left(\frac12\right)^x\).
Since:
\[ \frac14=\left(\frac12\right)^2 \]
we have:
\[ \left(\frac14\right)^x=\left(\frac12\right)^{2x} \]
Let:
\[ t=\left(\frac12\right)^x \]
with:
\[ t>0 \]
Then:
\[ \left(\frac12\right)^{2x}=t^2 \]
The inequality becomes:
\[ t^2-5t+4\le 0 \]
Factoring:
\[ t^2-5t+4=(t-1)(t-4) \]
So:
\[ (t-1)(t-4)\le 0 \]
The product is less than or equal to zero between the two roots:
\[ 1\le t\le 4 \]
Back-substituting \(t=\left(\frac12\right)^x\):
\[ 1\le \left(\frac12\right)^x\le 4 \]
Writing the bounds as powers of \(\frac12\):
\[ 1=\left(\frac12\right)^0,\qquad 4=\left(\frac12\right)^{-2} \]
Since the base \(\frac12\) satisfies \(0<\frac12<1\), the exponential function is decreasing, so the order of the exponents is reversed when we compare them.
We solve the two conditions separately.
From:
\[ \left(\frac12\right)^x\ge 1=\left(\frac12\right)^0 \]
the decreasing nature of the function gives:
\[ x\le 0 \]
From:
\[ \left(\frac12\right)^x\le 4=\left(\frac12\right)^{-2} \]
the decreasing nature of the function gives:
\[ x\ge -2 \]
Taking the intersection of the two conditions:
\[ -2\le x\le 0 \]
Therefore:
\[ S=[-2,0] \]
Exercise 18 — level ★★★★☆
Solve:
\[ \begin{cases} 2^x>4\\ 3^{x-1}\le 9 \end{cases} \]
Answer
\[ S=(2,3] \]
Solution
We solve each inequality separately.
First inequality:
\[ 2^x>4 \]
Since \(4=2^2\), we have:
\[ 2^x>2^2 \]
Since \(2>1\), we obtain:
\[ x>2 \]
Second inequality:
\[ 3^{x-1}\le 9 \]
Since \(9=3^2\), we obtain:
\[ 3^{x-1}\le 3^2 \]
Since \(3>1\), we compare exponents:
\[ x-1\le 2 \]
Hence:
\[ x\le 3 \]
Taking the intersection of both conditions:
\[ x>2 \quad \text{and} \quad x\le 3 \]
We obtain:
\[ 2
Therefore:
\[ S=(2,3] \]
Exercise 19 — level ★★★★★
Solve:
\[ \frac{2^{2x}-5\cdot 2^x+4}{2^x-2}\ge 0 \]
Answer
\[ S=[0,1)\cup[2,+\infty) \]
Solution
Let:
\[ t=2^x \]
Since \(2^x>0\), we have:
\[ t>0 \]
Moreover:
\[ 2^{2x}=(2^x)^2=t^2 \]
The inequality becomes:
\[ \frac{t^2-5t+4}{t-2}\ge 0 \]
Factoring the numerator:
\[ t^2-5t+4=(t-1)(t-4) \]
So:
\[ \frac{(t-1)(t-4)}{t-2}\ge 0 \]
The critical values are:
\[ t=1,\qquad t=2,\qquad t=4 \]
The value \(t=2\) makes the denominator zero and must therefore be excluded.
We analyze the sign on each subinterval for \(t>0\):
\[ \begin{array}{c|cccc} t & (0,1) & (1,2) & (2,4) & (4,+\infty)\\ \hline t-1 & - & + & + & +\\ t-4 & - & - & - & +\\ t-2 & - & - & + & +\\ \hline \dfrac{(t-1)(t-4)}{t-2} & - & + & - & + \end{array} \]
Since we need the expression to be greater than or equal to zero, we take the intervals where the sign is positive and include the zeros of the numerator:
\[ 1\le t<2 \quad \text{or} \quad t\ge 4 \]
The value \(t=2\) remains excluded because it makes the denominator zero.
Back-substituting \(t=2^x\):
\[ 1\le 2^x<2 \quad \text{or} \quad 2^x\ge 4 \]
Writing:
\[ 1=2^0,\qquad 2=2^1,\qquad 4=2^2 \]
and using the fact that \(2>1\), we obtain:
\[ 0\le x<1 \quad \text{or} \quad x\ge 2 \]
Therefore:
\[ S=[0,1)\cup[2,+\infty) \]
Exercise 20 — level ★★★★★
Solve:
\[ 4^x-3\cdot 2^{x+1}+8<0 \]
Answer
\[ S=(1,2) \]
Solution
We rewrite everything in terms of \(2^x\).
Since:
\[ 4^x=(2^2)^x=2^{2x}=(2^x)^2 \]
and:
\[ 2^{x+1}=2\cdot 2^x \]
let:
\[ t=2^x \]
with:
\[ t>0 \]
The inequality becomes:
\[ t^2-3\cdot 2t+8<0 \]
that is:
\[ t^2-6t+8<0 \]
Factoring:
\[ t^2-6t+8=(t-2)(t-4) \]
So:
\[ (t-2)(t-4)<0 \]
The product is negative between the two roots:
\[ 2
Back-substituting \(t=2^x\):
\[ 2<2^x<4 \]
Writing:
\[ 2=2^1,\qquad 4=2^2 \]
and using the fact that \(2>1\), we obtain:
\[ 1
Therefore:
\[ S=(1,2) \]