A progressive collection of step-by-step solved exercises to learn how to determine the domain, correctly apply the properties of logarithms, and verify the solutions obtained.
Exercise 1 — level ★☆☆☆☆
\[ \log_2 x = 3 \]
Answer
\[ x=8 \]
Solution
Domain
The argument of the logarithm must be positive: \[ x>0 \]
Solving the equation
We switch from logarithmic form to exponential form: \[ \log_2 x=3 \iff x=2^3 \]
Hence: \[ x=8 \]
Check
The solution lies in the domain, since: \[ 8>0 \]
Moreover: \[ \log_2 8=3 \]
Therefore: \[ \boxed{x=8} \]
Exercise 2 — level ★☆☆☆☆
\[ \log_3(x-1)=2 \]
Answer
\[ x=10 \]
Solution
Domain
The argument of the logarithm must be positive: \[ x-1>0 \] \[ x>1 \]
Solving the equation
We rewrite the logarithmic equation in exponential form: \[ \log_3(x-1)=2 \iff x-1=3^2 \]
Hence: \[ x-1=9 \] \[ x=10 \]
Check
The solution lies in the domain, since: \[ 10>1 \]
Moreover: \[ \log_3(10-1)=\log_3 9=2 \]
Therefore: \[ \boxed{x=10} \]
Exercise 3 — level ★☆☆☆☆
\[ \log_{10}(2x)=1 \]
Answer
\[ x=5 \]
Solution
Domain
The argument of the logarithm must be positive: \[ 2x>0 \] \[ x>0 \]
Solving the equation
We switch from logarithmic form to exponential form: \[ \log_{10}(2x)=1 \iff 2x=10^1 \]
Hence: \[ 2x=10 \] \[ x=5 \]
Check
The solution lies in the domain, since: \[ 5>0 \]
Moreover: \[ \log_{10}(2\cdot 5)=\log_{10}10=1 \]
Therefore: \[ \boxed{x=5} \]
Exercise 4 — level ★★☆☆☆
\[ \log_2(x+3)=\log_2 7 \]
Answer
\[ x=4 \]
Solution
Domain
The argument of the logarithm must be positive: \[ x+3>0 \] \[ x>-3 \]
Solving the equation
The two logarithms share the same base. Since the logarithmic function is one-to-one, we can equate the arguments: \[ x+3=7 \]
Solving: \[ x=4 \]
Check
The solution lies in the domain, since: \[ 4>-3 \]
Moreover: \[ \log_2(4+3)=\log_2 7 \]
Therefore: \[ \boxed{x=4} \]
Exercise 5 — level ★★☆☆☆
\[ \log_5(3x-1)=\log_5(x+7) \]
Answer
\[ x=4 \]
Solution
Domain
The arguments of the logarithms must be positive: \[ 3x-1>0 \] \[ x>\frac{1}{3} \] and \[ x+7>0 \] \[ x>-7 \]
The domain is therefore: \[ x>\frac{1}{3} \]
Solving the equation
The logarithms share the same base. We equate the arguments: \[ 3x-1=x+7 \]
Solving: \[ 2x=8 \] \[ x=4 \]
Check
The solution lies in the domain, since: \[ 4>\frac{1}{3} \]
Moreover: \[ 3\cdot 4-1=11 \] \[ 4+7=11 \] so the two logarithms have indeed the same argument.
Therefore: \[ \boxed{x=4} \]
Exercise 6 — level ★★☆☆☆
\[ \log_2 x+\log_2 4=5 \]
Answer
\[ x=8 \]
Solution
Domain
The only argument containing the unknown must be positive: \[ x>0 \]
Solving the equation
We compute the known logarithm: \[ \log_2 4=2 \]
The equation becomes: \[ \log_2 x+2=5 \]
Hence: \[ \log_2 x=3 \]
We switch to exponential form: \[ x=2^3=8 \]
Check
The solution lies in the domain, since: \[ 8>0 \]
Moreover: \[ \log_2 8+\log_2 4=3+2=5 \]
Therefore: \[ \boxed{x=8} \]
Exercise 7 — level ★★☆☆☆
\[ \log_3 x+\log_3(x-2)=1 \]
Answer
\[ x=3 \]
Solution
Domain
The arguments must be positive: \[ x>0 \] \[ x-2>0 \]
Hence: \[ x>2 \]
Solving the equation
We use the product rule for logarithms: \[ \log_a A+\log_a B=\log_a(AB) \]
We obtain: \[ \log_3[x(x-2)]=1 \]
We switch to exponential form: \[ x(x-2)=3^1 \] \[ x(x-2)=3 \]
Expanding: \[ x^2-2x=3 \] \[ x^2-2x-3=0 \]
Factoring: \[ (x-3)(x+1)=0 \]
Hence: \[ x=3 \quad \text{or} \quad x=-1 \]
Check
The domain requires \(x>2\). Therefore: \[ x=-1 \] is rejected, while \[ x=3 \] is acceptable.
Indeed: \[ \log_3 3+\log_3(3-2)=1+0=1 \]
Therefore: \[ \boxed{x=3} \]
Exercise 8 — level ★★☆☆☆
\[ \log_2(x+1)+\log_2(x-1)=3 \]
Answer
\[ x=3 \]
Solution
Domain
The arguments must be positive: \[ x+1>0 \Rightarrow x>-1 \] \[ x-1>0 \Rightarrow x>1 \]
Hence: \[ x>1 \]
Solving the equation
We apply the product rule: \[ \log_2(x+1)+\log_2(x-1)=\log_2[(x+1)(x-1)] \]
Hence: \[ \log_2(x^2-1)=3 \]
We switch to exponential form: \[ x^2-1=2^3 \] \[ x^2-1=8 \] \[ x^2=9 \]
Therefore: \[ x=-3 \quad \text{or} \quad x=3 \]
Check
The domain requires \(x>1\), so \(x=-3\) is rejected.
For \(x=3\): \[ \log_2(3+1)+\log_2(3-1)=\log_2 4+\log_2 2=2+1=3 \]
Therefore: \[ \boxed{x=3} \]
Exercise 9 — level ★★★☆☆
\[ \log_4(x+6)-\log_4 x=1 \]
Answer
\[ x=2 \]
Solution
Domain
The arguments must be positive: \[ x+6>0 \] \[ x>0 \]
The more restrictive condition is: \[ x>0 \]
Solving the equation
We apply the quotient rule: \[ \log_a A-\log_a B=\log_a\left(\frac{A}{B}\right) \]
We obtain: \[ \log_4\left(\frac{x+6}{x}\right)=1 \]
We switch to exponential form: \[ \frac{x+6}{x}=4^1 \] \[ \frac{x+6}{x}=4 \]
On the domain \(x>0\), we may multiply both sides by \(x\): \[ x+6=4x \]
Solving: \[ 6=3x \] \[ x=2 \]
Check
The solution lies in the domain, since: \[ 2>0 \]
Moreover: \[ \log_4(2+6)-\log_4 2=\log_4 8-\log_4 2=\log_4 4=1 \]
Therefore: \[ \boxed{x=2} \]
Exercise 10 — level ★★★☆☆
\[ \log_2(x+2)-\log_2(x-1)=2 \]
Answer
\[ x=2 \]
Solution
Domain
The arguments must be positive: \[ x+2>0 \Rightarrow x>-2 \] \[ x-1>0 \Rightarrow x>1 \]
Hence: \[ x>1 \]
Solving the equation
We apply the quotient rule: \[ \log_2(x+2)-\log_2(x-1)=\log_2\left(\frac{x+2}{x-1}\right) \]
Hence: \[ \log_2\left(\frac{x+2}{x-1}\right)=2 \]
We switch to exponential form: \[ \frac{x+2}{x-1}=2^2 \] \[ \frac{x+2}{x-1}=4 \]
On the domain \(x>1\), we have \(x-1>0\), so we may multiply both sides by \(x-1\): \[ x+2=4(x-1) \]
Expanding: \[ x+2=4x-4 \] \[ 6=3x \] \[ x=2 \]
Check
The solution lies in the domain, since: \[ 2>1 \]
Moreover: \[ \log_2(2+2)-\log_2(2-1)=\log_2 4-\log_2 1=2-0=2 \]
Therefore: \[ \boxed{x=2} \]
Exercise 11 — level ★★★☆☆
\[ \log_3(x^2-4)=2 \]
Answer
\[ x=-\sqrt{13} \quad \text{or} \quad x=\sqrt{13} \]
Solution
Domain
The argument of the logarithm must be positive: \[ x^2-4>0 \]
Factoring: \[ x^2-4=(x-2)(x+2) \]
Hence: \[ (x-2)(x+2)>0 \]
The product is positive when both factors share the same sign: \[ x<-2 \quad \text{or} \quad x>2 \]
Solving the equation
We switch from logarithmic form to exponential form: \[ \log_3(x^2-4)=2 \iff x^2-4=3^2 \]
Hence: \[ x^2-4=9 \] \[ x^2=13 \]
Therefore: \[ x=-\sqrt{13} \quad \text{or} \quad x=\sqrt{13} \]
Check
Both solutions lie in the domain, since: \[ -\sqrt{13}<-2 \] and \[ \sqrt{13}>2 \]
Moreover, in both cases: \[ x^2=13 \] hence: \[ \log_3(x^2-4)=\log_3(13-4)=\log_3 9=2 \]
Therefore: \[ \boxed{x=-\sqrt{13} \quad \text{or} \quad x=\sqrt{13}} \]
Exercise 12 — level ★★★☆☆
\[ \log_2(x^2-5x+6)=1 \]
Answer
\[ x=1 \quad \text{or} \quad x=4 \]
Solution
Domain
The argument of the logarithm must be positive: \[ x^2-5x+6>0 \]
Factoring the trinomial: \[ x^2-5x+6=(x-2)(x-3) \]
Hence: \[ (x-2)(x-3)>0 \]
The product is positive outside the interval between the two roots: \[ x<2 \quad \text{or} \quad x>3 \]
Solving the equation
We switch to exponential form: \[ \log_2(x^2-5x+6)=1 \iff x^2-5x+6=2^1 \]
Hence: \[ x^2-5x+6=2 \]
Moving everything to the left-hand side: \[ x^2-5x+4=0 \]
Factoring: \[ (x-1)(x-4)=0 \]
Hence: \[ x=1 \quad \text{or} \quad x=4 \]
Check
Both solutions lie in the domain: \[ 1<2 \] and \[ 4>3 \]
We check in the original equation.
For \(x=1\): \[ \log_2(1^2-5\cdot 1+6)=\log_2 2=1 \]
For \(x=4\): \[ \log_2(4^2-5\cdot 4+6)=\log_2 2=1 \]
Therefore: \[ \boxed{x=1 \quad \text{or} \quad x=4} \]
Exercise 13 — level ★★★★☆
\[ \log_2 x+\log_2(x+2)=\log_2 15 \]
Answer
\[ x=3 \]
Solution
Domain
The arguments of the logarithms must be positive: \[ x>0 \] and \[ x+2>0 \]
The more restrictive condition is: \[ x>0 \]
Solving the equation
We apply the product rule for logarithms: \[ \log_2 x+\log_2(x+2)=\log_2[x(x+2)] \]
The equation becomes: \[ \log_2[x(x+2)]=\log_2 15 \]
Since the logarithms share the same base, we equate the arguments: \[ x(x+2)=15 \]
Expanding: \[ x^2+2x=15 \] \[ x^2+2x-15=0 \]
Factoring: \[ (x+5)(x-3)=0 \]
Hence: \[ x=-5 \quad \text{or} \quad x=3 \]
Check
The domain requires \(x>0\), so \(x=-5\) is rejected.
For \(x=3\): \[ \log_2 3+\log_2(3+2)=\log_2 3+\log_2 5=\log_2 15 \]
Therefore: \[ \boxed{x=3} \]
Exercise 14 — level ★★★★☆
\[ \log_3(x+1)+\log_3(x+3)=2 \]
Answer
\[ x=-2+\sqrt{10} \]
Solution
Domain
The arguments must be positive: \[ x+1>0 \Rightarrow x>-1 \] \[ x+3>0 \Rightarrow x>-3 \]
The domain is therefore: \[ x>-1 \]
Solving the equation
We apply the product rule: \[ \log_3(x+1)+\log_3(x+3)=\log_3[(x+1)(x+3)] \]
We obtain: \[ \log_3[(x+1)(x+3)]=2 \]
We switch to exponential form: \[ (x+1)(x+3)=3^2 \] \[ (x+1)(x+3)=9 \]
Expanding: \[ x^2+4x+3=9 \] \[ x^2+4x-6=0 \]
Applying the quadratic formula: \[ x=\frac{-4\pm\sqrt{4^2-4\cdot 1\cdot(-6)}}{2} \] \[ x=\frac{-4\pm\sqrt{16+24}}{2} \] \[ x=\frac{-4\pm\sqrt{40}}{2} \] \[ x=\frac{-4\pm 2\sqrt{10}}{2} \] \[ x=-2\pm\sqrt{10} \]
Check
The domain requires \(x>-1\).
The solution \[ x=-2-\sqrt{10} \] is less than \(-1\), so it is rejected.
The solution \[ x=-2+\sqrt{10} \] is greater than \(-1\), so it is acceptable.
Moreover, by construction of the solving step: \[ (x+1)(x+3)=9 \] hence: \[ \log_3(x+1)+\log_3(x+3)=\log_3 9=2 \]
Therefore: \[ \boxed{x=-2+\sqrt{10}} \]
Exercise 15 — level ★★★★☆
\[ \log_2(x+4)-\log_2(x-2)=1 \]
Answer
\[ x=8 \]
Solution
Domain
The arguments must be positive: \[ x+4>0 \Rightarrow x>-4 \] \[ x-2>0 \Rightarrow x>2 \]
Hence: \[ x>2 \]
Solving the equation
We apply the quotient rule: \[ \log_2(x+4)-\log_2(x-2)=\log_2\left(\frac{x+4}{x-2}\right) \]
Hence: \[ \log_2\left(\frac{x+4}{x-2}\right)=1 \]
We switch to exponential form: \[ \frac{x+4}{x-2}=2^1 \] \[ \frac{x+4}{x-2}=2 \]
On the domain \(x>2\), we have \(x-2>0\). We may therefore multiply both sides by \(x-2\): \[ x+4=2(x-2) \]
Expanding: \[ x+4=2x-4 \] \[ x=8 \]
Check
The solution lies in the domain, since: \[ 8>2 \]
Moreover: \[ \log_2(8+4)-\log_2(8-2)=\log_2 12-\log_2 6=\log_2 2=1 \]
Therefore: \[ \boxed{x=8} \]
Exercise 16 — level ★★★★☆
\[ \log x+\log(x-9)=1 \]
Answer
\[ x=10 \]
Solution
Domain
When the base is not specified, we mean the common (decimal) logarithm: \[ \log x=\log_{10}x \]
The arguments must be positive: \[ x>0 \] \[ x-9>0 \Rightarrow x>9 \]
The domain is therefore: \[ x>9 \]
Solving the equation
We apply the product rule: \[ \log x+\log(x-9)=\log[x(x-9)] \]
We obtain: \[ \log[x(x-9)]=1 \]
We switch to exponential form in base \(10\): \[ x(x-9)=10^1 \] \[ x(x-9)=10 \]
Expanding: \[ x^2-9x=10 \] \[ x^2-9x-10=0 \]
Factoring: \[ (x-10)(x+1)=0 \]
Hence: \[ x=10 \quad \text{or} \quad x=-1 \]
Check
The domain requires \(x>9\), so \(x=-1\) is rejected.
For \(x=10\): \[ \log 10+\log(10-9)=\log 10+\log 1=1+0=1 \]
Therefore: \[ \boxed{x=10} \]
Exercise 17 — level ★★★★★
\[ \log_2(x-1)+\log_2(x+1)=\log_2(2x+6) \]
Answer
\[ x=1+2\sqrt{2} \]
Solution
Domain
The arguments must be positive: \[ x-1>0 \Rightarrow x>1 \] \[ x+1>0 \Rightarrow x>-1 \] \[ 2x+6>0 \Rightarrow x>-3 \]
The domain is therefore: \[ x>1 \]
Solving the equation
We apply the product rule on the left-hand side: \[ \log_2(x-1)+\log_2(x+1)=\log_2[(x-1)(x+1)] \]
The equation becomes: \[ \log_2[(x-1)(x+1)]=\log_2(2x+6) \]
Since the logarithms share the same base, we equate the arguments: \[ (x-1)(x+1)=2x+6 \]
Expanding: \[ x^2-1=2x+6 \] \[ x^2-2x-7=0 \]
Applying the quadratic formula: \[ x=\frac{2\pm\sqrt{(-2)^2-4\cdot 1\cdot(-7)}}{2} \] \[ x=\frac{2\pm\sqrt{4+28}}{2} \] \[ x=\frac{2\pm\sqrt{32}}{2} \] \[ x=1\pm 2\sqrt{2} \]
Check
The domain requires \(x>1\).
The solution \[ x=1-2\sqrt{2} \] is less than \(1\), so it is rejected.
The solution \[ x=1+2\sqrt{2} \] is greater than \(1\), so it is acceptable.
Therefore: \[ \boxed{x=1+2\sqrt{2}} \]
Exercise 18 — level ★★★★★
\[ \log_3 x+\log_3(x+6)=\log_3(7x+18) \]
Answer
\[ x=\frac{1+\sqrt{73}}{2} \]
Solution
Domain
The arguments must be positive: \[ x>0 \] \[ x+6>0 \Rightarrow x>-6 \] \[ 7x+18>0 \Rightarrow x>-\frac{18}{7} \]
The more restrictive condition is: \[ x>0 \]
Solving the equation
We apply the product rule on the left-hand side: \[ \log_3 x+\log_3(x+6)=\log_3[x(x+6)] \]
Hence: \[ \log_3[x(x+6)]=\log_3(7x+18) \]
We equate the arguments: \[ x(x+6)=7x+18 \]
Expanding: \[ x^2+6x=7x+18 \] \[ x^2-x-18=0 \]
Applying the quadratic formula: \[ x=\frac{1\pm\sqrt{(-1)^2-4\cdot 1\cdot(-18)}}{2} \] \[ x=\frac{1\pm\sqrt{1+72}}{2} \] \[ x=\frac{1\pm\sqrt{73}}{2} \]
Check
The domain requires \(x>0\).
The solution \[ x=\frac{1-\sqrt{73}}{2} \] is negative, so it is rejected.
The solution \[ x=\frac{1+\sqrt{73}}{2} \] is positive, so it is acceptable.
Therefore: \[ \boxed{x=\frac{1+\sqrt{73}}{2}} \]
Exercise 19 — level ★★★★★
\[ \log_2(x+2)+\log_2(x-2)=\log_2 12 \]
Answer
\[ x=4 \]
Solution
Domain
The arguments must be positive: \[ x+2>0 \Rightarrow x>-2 \] \[ x-2>0 \Rightarrow x>2 \]
Hence: \[ x>2 \]
Solving the equation
We apply the product rule: \[ \log_2(x+2)+\log_2(x-2)=\log_2[(x+2)(x-2)] \]
The equation becomes: \[ \log_2[(x+2)(x-2)]=\log_2 12 \]
We equate the arguments: \[ (x+2)(x-2)=12 \]
Using the difference of squares: \[ x^2-4=12 \] \[ x^2=16 \]
Hence: \[ x=-4 \quad \text{or} \quad x=4 \]
Check
The domain requires \(x>2\), so \(x=-4\) is rejected.
For \(x=4\): \[ \log_2(4+2)+\log_2(4-2)=\log_2 6+\log_2 2=\log_2 12 \]
Therefore: \[ \boxed{x=4} \]
Exercise 20 — level ★★★★★
\[ \log_3(x-2)+\log_3(x+2)=2 \]
Answer
\[ x=\sqrt{13} \]
Solution
Domain
The arguments must be positive: \[ x-2>0 \Rightarrow x>2 \] \[ x+2>0 \Rightarrow x>-2 \]
The domain is therefore: \[ x>2 \]
Solving the equation
We apply the product rule: \[ \log_3(x-2)+\log_3(x+2)=\log_3[(x-2)(x+2)] \]
We obtain: \[ \log_3[(x-2)(x+2)]=2 \]
We switch to exponential form: \[ (x-2)(x+2)=3^2 \] \[ (x-2)(x+2)=9 \]
Using the difference of squares: \[ x^2-4=9 \] \[ x^2=13 \]
Hence: \[ x=-\sqrt{13} \quad \text{or} \quad x=\sqrt{13} \]
Check
The domain requires \(x>2\), so \(x=-\sqrt{13}\) is rejected.
Since: \[ \sqrt{13}>2 \] the solution \(x=\sqrt{13}\) is acceptable.
Moreover: \[ (\sqrt{13}-2)(\sqrt{13}+2)=13-4=9 \] hence: \[ \log_3(\sqrt{13}-2)+\log_3(\sqrt{13}+2)=\log_3 9=2 \]
Therefore: \[ \boxed{x=\sqrt{13}} \]