1B - Solving logarithmic equations

Why can algebra produce two candidates when a logarithmic equation has only one solution? In 1B you will isolate or combine logarithms, reverse them into powers and check the answers in the original equation. The positive-argument conditions will decide which algebraic candidates really work.

Isolating a logarithm

An equation with one logarithm becomes an ordinary algebraic equation once that logarithm is isolated.

For instance, $\log_2(3x-1)=4$ says that the whole argument $3x-1$ , rather than $x$ alone, equals $2^4$ . The logarithm's value is an exponent, not the argument's value.

Reversing a logarithmic equation

For a valid base $b>1$ and positive argument $F(x)$ ,

\log_b(F(x))=c\quad\Longleftrightarrow\quad F(x)=b^c.

First isolate the logarithm using ordinary algebra. Keep brackets around its complete argument.

Write any restrictions from the original equation before rearranging. For a logarithm such as $\log_2(3x-1)$ , the condition is $3x-1>0$ , so $x>\frac13$ .

Worked example 1: an unknown argument

Question. Solve $\log_2(3x-1)=4$ .

Plan. Reverse the logarithm, solve the linear equation and check the original argument.

\begin{aligned} 3x-1&=2^4 &&\text{logarithm gives the exponent}\\ 3x-1&=16\\ 3x&=17\\ x&=\frac{17}{3} \end{aligned}

Answer. $x=\frac{17}{3}$ .

Check. Substitution gives $3(17/3)-1=16>0$ , and $\log_2 16=4$ , as required.

The unknown can also be the base. Isolating the logarithm still comes first; afterwards the unknown appears as the base of a power.

Worked example 2: isolating a logarithm with an unknown base

Question. Solve $2\log_b 16=4$ , where $b>1$ .

Plan. Divide by $2$ , change to exponential form and use the restriction on the base.

\begin{aligned} \log_b 16&=2 &&\text{divide both sides by }2\\ b^2&=16 &&\text{reverse the logarithm}\\ b&=4 &&\text{retain the positive base greater than }1 \end{aligned}

The algebraic square-root candidate $-4$ is not a valid real logarithm base.

Answer. $b=4$ .

Check. $4^2=16$ , so $2\log_4 16=2\times2=4$ .

If an equation has an added constant outside the logarithm, move that constant before reversing the log. If the log has a multiplier, divide by it. Do not apply the exponential conversion to only part of one side.

Combining before solving

When several same-base logarithms occur, combine them into one logarithm before converting to powers.

The product, quotient and power laws are useful because they change the form of an equation without changing its values on the original domain. A sum gives a product of arguments, a difference gives a quotient, and a multiplier gives a power.

Worked example 3: combining a sum

Question. Solve $\log_2(x-1)+\log_2 2=3$ .

Plan. Require $x>1$ , combine the sum and reverse the resulting logarithm.

\begin{aligned} \log_2(2(x-1))&=3 &&\text{product law}\\ 2(x-1)&=2^3 &&\text{reverse the logarithm}\\ 2x-2&=8\\ x&=5 \end{aligned}

Answer. $x=5$ .

Check. In the original equation, $\log_2 4+\log_2 2=2+1=3$ ; both arguments are positive. Alternatively, evaluating $\log_2 2=1$ first gives $\log_2(x-1)=2$ , with the same solution.

Now replace the sum by a difference, with the unknown in both arguments. The order of the quotient matters.

Worked example 4: combining a difference

Question. Solve $\log_3(2x+1)-\log_3(x-1)=1$ .

Plan. Use the original domain $x>1$ , combine with the quotient law and solve.

\begin{aligned} \log_3\!\left(\frac{2x+1}{x-1}\right)&=1 &&\text{first argument divided by second}\\ \frac{2x+1}{x-1}&=3\\ 2x+1&=3(x-1) &&\text{multiply by the nonzero denominator}\\ 2x+1&=3x-3\\ x&=4 \end{aligned}

Answer. $x=4$ .

Check. The original arguments are $9$ and $3$ , so $\log_3 9-\log_3 3=2-1=1$ .

Two logarithms of the same base are equal only when their positive arguments are equal. This follows because each output on the increasing logarithmic graph occurs once. Thus an equation such as $\log_3 A=\log_3 B$ , with $A,B>0$ , can be solved by setting $A=B$ .

Checking logarithmic solutions

Algebraic solutions of a combined equation are candidates until each original logarithm has been checked.

For $\log_b(x-1)+\log_b(x+1)$ , both $x-1>0$ and $x+1>0$ are needed, so $x>1$ . The combined argument $(x-1)(x+1)$ can also be positive when both factors are negative. Those inputs do not belong to the original equation.

Worked example 5: retaining one of two candidates

Question. Solve $\log_3(x-1)+\log_3(x+1)=1$ .

Plan. Record $x>1$ , combine and solve the square equation, then select valid candidates.

\begin{aligned} \log_3((x-1)(x+1))&=1\\ (x-1)(x+1)&=3\\ x^2-1&=3\\ x^2&=4\\ x&=2\quad\text{or}\quad x=-2 \end{aligned}

Only $2$ satisfies $x>1$ . At $x=-2$ , the original arguments are $-3$ and $-1$ , so neither original logarithm has a real value.

Answer. $x=2$ .

Check. At $x=2$ , the original left side is $\log_3 1+\log_3 3=0+1=1$ .

Sometimes every algebraic candidate fails. A logarithmic equation can have no real solution.

Worked example 6: rejecting the only candidate

Question. Solve $\log_2(x-3)-\log_2(x-5)=-1$ .

Plan. Keep the original domain $x>5$ , then combine and solve.

\begin{aligned} \log_2\!\left(\frac{x-3}{x-5}\right)&=-1\\ \frac{x-3}{x-5}&=\frac12\\ 2(x-3)&=x-5\\ x&=1 \end{aligned}

At $x=1$ , the original arguments are $-2$ and $-4$ , so the candidate is invalid.

Answer. There is no real solution.

Check. For every permitted $x>5$ , the positive numerator $x-3$ is greater than the denominator $x-5$ . Their ratio exceeds $1$ , so its base-two logarithm is positive and cannot equal $-1$ .

The correct restriction depends on the original expression. For $\log_2(x^2)$ , positivity of the whole argument requires only $x\ne0$ . For $2\log_2 x$ , it requires $x>0$ . These expressions agree on positive inputs, but their original domains are different.

Key idea

Check the argument of every original logarithm, including any brackets or powers. A positive product or quotient alone is not enough to validate the original separate logarithms.

Use the original arguments to decide whether the next equation has a solution.

Quick Recap

Isolate or combine, convert to powers, solve, then check every original logarithm. The opening issue occurs because a combined product can be positive when its separate factors are negative: the resulting algebra may admit candidates that the original equation excludes.

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1B - Solving logarithmic equations

Isolating a logarithm

Combining before solving

Checking logarithmic solutions

Quick Recap

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