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Solving Exponential and Logarithmic Equations: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A quotient equation

Solve $\frac{8^x}{2^{x+1}}=4^{3-x}$ for real $x$ .
1. Hint 1
  Write every exponential using base $2$ .
2. Hint 2
  The denominator is positive, so dividing powers subtracts their exponents safely.
Answer

$x=\frac74$ .

Full solution

The positive denominator permits rewriting the left side as $2^{2x-1}$ ; the right side is $2^{6-2x}$ .
One-to-oneness gives
$2x-1=6-2x$
Thus $4x=7$ and $x=\frac74$ .
At that value both exponents are $\frac52$ , verifying the equation.

Answer

$x=\frac74$ .

Key idea

For $b>0$ and $b\ne1$ , $b^u=b^v$ is equivalent to $u=v$ , which turns an exponential equation into an equality of exponents.
Problem 1: Mark complete
Problem 2 A square-root argument

Solve $\log_3(\sqrt{x+1})=1$ for real $x$ .
1. Hint 1
  The logarithm requires the square root to be positive.
2. Hint 2
  Translate the logarithm into an equation for the square root.
Answer

$x=8$ .

Full solution

The domain is $x>-1$ .
The logarithm definition gives
$\sqrt{x+1}=3.$
Squaring yields
$x+1=9$
so $x=8$ .
Its argument is $\sqrt9=3>0$ , and $\log_3 3=1$ , so it checks.

Answer

$x=8$ .

Key idea

Translate a logarithm first, then solve for the expression inside its argument.
Problem 2: Mark complete
Problem 3 Two logarithm bases

Solve $\log_2(x+2)=1+\log_4(x+2)$ for real $x$ .
1. Hint 1
  The two logarithms use the same positive argument.
2. Hint 2
  Convert the base-4 logarithm to base $2$ .
Answer

$x=2$ .

Full solution

The domain is $x>-2$ .
Put $u=\log_2(x+2)$ , so
$\log_4(x+2)=u/2.$
The equation becomes
$u=1+\frac u2$
giving $u=2$ .

Thus $x+2=2^2=4$ , so $x=2$ .
The original sides are $2$ and $1+1$ , respectively.

Answer

$x=2$ .

Key idea

Changing bases can turn an equation with two logarithms into one linear equation.
Problem 3: Mark complete
Problem 4 A crossing time

Two positive readings follow $A(t)=12\cdot1.5^t$ and $B(t)=30\cdot1.2^t$ for $t\ge0$ , with $t$ in minutes. Find when the readings agree, both exactly and to the nearest hundredth of a minute.
1. Hint 1
  Divide the readings to isolate one exponential factor.
2. Hint 2
  Take logarithms after isolating a power of $1.5/1.2$ .
Answer

$t=\frac{\log_{10}(5/2)}{\log_{10}(5/4)}\approx4.11$ minutes.

Full solution

Both readings are positive, so division is legal.
Equality reduces to
$1.25^t=2.5.$
Taking base-10 logarithms gives
$t\log_{10}(5/4)=\log_{10}(5/2)$

The denominator $\log_{10}(5/4)$ is positive.
Division gives the exact expression stated and $t\approx4.11$ minutes.
Substitution into the ratio gives $A/B=1$ exactly at the unrounded time.

Answer

$t=\frac{\log_{10}(5/2)}{\log_{10}(5/4)}\approx4.11$ minutes.

Key idea

Comparing two exponential readings can reduce to solving one exponential equation.
Problem 4: Mark complete
Problem 5 A reciprocal pair

Find every real solution of $2^x+2^{-x}=\frac52$ .
1. Hint 1
  The two powers are reciprocals.
2. Hint 2
  Let $u=2^x>0$ and multiply the equation by $2u$ .
Answer

$x=-1$ or $x=1$ .

Full solution

Let $u=2^x>0$ .
Then $2^{-x}=1/u$ , so multiplying by $2u$ , which is nonzero, gives
$2u^2-5u+2=0$
This factors as
$(2u-1)(u-2)=0.$

The positive candidates are $u=\frac12$ and $u=2$ , giving $x=-1$ and $x=1$ .
Each yields the same original sum,
$\frac12+2=\frac52.$

Answer

$x=-1$ or $x=1$ .

Key idea

A reciprocal exponential pair can become a quadratic in one positive quantity.
Problem 5: Mark complete
Problem 6 A sum of logarithms

Solve $\log_2(x+1)+\log_2(5-x)=3$ for real $x$ .
1. Hint 1
  Both arguments must be positive before they can be combined.
2. Hint 2
  The product of the two arguments must equal $2^3$ .
Answer

$x=1$ or $x=3$ .

Full solution

The domain is $-1<x<5$ .
Combining gives $(x+1)(5-x)=8$ , so
$x^2-4x+3=0$
The candidates are $1$ and $3$ .

Both are in the domain.
At $x=1$ , the logarithms are $1$ and $2$ ; at $x=3$ , they are $2$ and $1$ .
Both sums are $3$ .

Answer

$x=1$ or $x=3$ .

Key idea

Two logarithm arguments can impose both a lower and an upper bound on possible solutions.
Problem 6: Mark complete
Problem 7 A candidate list

Solve $\log_7(x^2-11)-\log_7(x-3)=1$ for real $x$ . Report every algebraic candidate and explain any rejection.
1. Hint 1
  The quotient rule requires both original arguments to be positive.
2. Hint 2
  Combine, convert to exponential form, and check each candidate in the original equation.
Answer

$x=5$ ; the candidate $x=2$ is rejected.

Full solution

The original arguments require $x^2>11$ and $x>3$ , that is,
$x>\sqrt{11}.$
On that domain the quotient rule gives
$\log_7\frac{x^2-11}{x-3}=1$

So $\frac{x^2-11}{x-3}=7$ , which gives $x^2-7x+10=0$ and
$(x-2)(x-5)=0$
The candidates are $2$ and $5$ .

At $5$ the arguments are $14$ and $2$ , both positive, and $\log_7 14-\log_7 2=\log_7 7$ , which is $1$ .
At $2$ the arguments are $-7$ and $-1$ : their quotient is $7$ , but neither original logarithm is defined, so $2$ is rejected.
Combining into one logarithm enlarged the domain, and that is where $2$ came from.

Answer

$x=5$ ; the candidate $x=2$ is rejected.

Key idea

Combining logarithms can enlarge the domain, so each candidate is checked in the original equation.
Problem 7: Mark complete
Problem 8 Two candidate values

A student solving $9^x+6\cdot3^x=0$ substitutes $u=3^x$ and obtains $u=0$ or $u=-6$ . The student concludes there is no real solution. Is that conclusion correct? Explain.
1. Hint 1
  Check the range of the expression represented by $u$ .
2. Hint 2
  You can also examine the signs of the original terms.
Answer

Yes; there is no real solution.

Full solution

The substitution gives $u(u+6)=0$ , but $u=3^x$ is strictly positive for every real $x$ , so neither candidate is possible.
Directly,
$9^x+6\cdot3^x>0$
at every real input.
Thus the original left side cannot equal zero.

Answer

Yes; there is no real solution.

Key idea

A substitution representing a positive exponential excludes zero and negative substitution values.
Problem 8: Mark complete
Problem 9 A recorded decimal

A solver writes $x=1.252$ for the solution of $5^x+3=7\cdot5^{x-1}$ . Is this exact or rounded? Give the exact solution and its value to the nearest thousandth.
1. Hint 1
  Both powers are constant multiples of the same exponential.
2. Hint 2
  Rewrite $5^{x-1}=\frac{5^x}5$ , collect the terms, and keep an exact logarithmic form.
Answer

Rounded, not exact: $x=\log_5\frac{15}2\approx1.252$ .

Full solution

The equation is
$5^x+3=\frac75\cdot5^x.$
Subtracting $5^x$ gives
$3=\frac25\cdot5^x$
so $5^x=\frac{15}2$ and
$x=\log_5\frac{15}2.$

Since $\frac{15}2$ is not a rational power of $5$ , this logarithm is irrational; its value is $1.2519\ldots$ , which rounds to $1.252$ .
So $1.252$ is a correct rounding but not the exact solution.

Answer

Rounded, not exact: $x=\log_5\frac{15}2\approx1.252$ .

Key idea

Keep the exact logarithmic form, and label a decimal that has been rounded as an approximation.
Problem 9: Mark complete
Problem 10 A shared logarithm value

A student says that $\log_5(x^2+4)=\log_5(4x)$ has exactly one real solution, even though equating the arguments gives a quadratic. Decide whether the claim is correct and justify it.
1. Hint 1
  A quadratic can have a repeated root.
2. Hint 2
  Keep the condition $4x>0$ when solving the argument equation.
Answer

Yes; the only solution is $x=2$ .

Full solution

The right argument requires $x>0$ ; the left is positive for every real input.
Equating arguments gives $x^2+4=4x$ , or
$(x-2)^2=0$
Thus $x=2$ is the sole candidate.
Both arguments are then $8>0$ , so it is a genuine solution.

Answer

Yes; the only solution is $x=2$ .

Key idea

A quadratic equation in logarithm arguments may have one repeated root that satisfies the original domain.
Problem 10: Mark complete

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