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Solve literal logarithmic and indicial equations

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Solving literal logarithmic equations

Isolate the logarithm, convert it to exponential form, and then rearrange the named variable. Keep the base and multiplier restrictions with the exact solution

\[\log_a\left(mx+n\right)=r\quad\Rightarrow\quad x=\frac{a^r-n}{m}\]
  1. 1Isolate the complete argument In a logarithmic equation, first isolate the expression that contains the named variable \(\displaystyle \log_a\left(mx+n\right)=r\quad\Rightarrow\quad mx+n=a^r\)
  2. 2Use the inverse form Convert a logarithm to an exponential with the same base \(\displaystyle \log_a\left(u\right)=r\quad\Longleftrightarrow\quad u=a^r\)
  3. 3Rearrange the named variable Move the constant term and divide by the non-zero coefficient of the named variable \(\displaystyle mx+n=a^r\quad\Rightarrow\quad x=\frac{a^r-n}{m}\)
  4. 4State the restrictions The logarithm base must be positive and not 1, and a divisor must be non-zero \(\displaystyle a>0,\quad a\ne1,\quad m\ne0\)
Year 12, VCE Units 3 & 4, Chapter 4, 4 difficulty levels, No calculatorBuild a worksheet

Solving literal logarithmic equations

Isolate the logarithm, convert it to exponential form, and then rearrange the named variable. Keep the base and multiplier restrictions with the exact solution

\[\log_a\left(mx+n\right)=r\quad\Rightarrow\quad x=\frac{a^r-n}{m}\]
  1. Isolate the complete argument In a logarithmic equation, first isolate the expression that contains the named variable \(\displaystyle \log_a\left(mx+n\right)=r\quad\Rightarrow\quad mx+n=a^r\)
  2. Use the inverse form Convert a logarithm to an exponential with the same base \(\displaystyle \log_a\left(u\right)=r\quad\Longleftrightarrow\quad u=a^r\)
  3. Rearrange the named variable Move the constant term and divide by the non-zero coefficient of the named variable \(\displaystyle mx+n=a^r\quad\Rightarrow\quad x=\frac{a^r-n}{m}\)
  4. State the restrictions The logarithm base must be positive and not 1, and a divisor must be non-zero \(\displaystyle a>0,\quad a\ne1,\quad m\ne0\)

Worked example

Solve the literal logarithmic equation below for x and state the restrictions on the parameters

\[\log_a\left(2mx-3n\right)=p\]
  1. \(\displaystyle 2mx-3n=a^p\)Convert the logarithmic equation to exponential form with the same base a.
  2. \(\displaystyle 2mx=a^p+3n\)Add 3n to both sides to isolate the term containing x.
  3. \(\displaystyle x=\frac{a^p+3n}{2m}\)Divide by 2m to solve for x.
  4. \(\displaystyle a>0,\quad a\ne1,\quad m\ne0,\quad 2mx-3n>0\)The base, divisor and original logarithm argument must satisfy their restrictions

Common mistakes

Solving for the wrong letter
✕\(\displaystyle \log_a\left(2mx-3n\right)=p\quad\Rightarrow\quad m=\frac{a^p+3n}{2x}\)✓\(\displaystyle \log_a\left(2mx-3n\right)=p\quad\Rightarrow\quad x=\frac{a^p+3n}{2m}\)
The named variable is x, so rearrangement must isolate x rather than a parameter
Failing to use the exponential inverse
✕\(\displaystyle \log_a\left(mx+n\right)=r\quad\Rightarrow\quad mx+n=r\)✓\(\displaystyle \log_a\left(mx+n\right)=r\quad\Rightarrow\quad mx+n=a^r\)
The inverse of log base a is raising a to the logarithm value, not leaving the value unchanged
Dropping the multiplier restriction
✕\(\displaystyle \log_a\left(mx+n\right)=r\quad\Rightarrow\quad a>0,\quad a\ne1\)✓\(\displaystyle \log_a\left(mx+n\right)=r\quad\Rightarrow\quad a>0,\quad a\ne1,\quad m\ne0\)
The solution divides by m, so m must be non-zero as well as the logarithm base being valid

Difficulty levels

introSolves one logarithmic or exponential equation with an integer coefficient and one restriction
standardSolves a two-parameter equation with a rational coefficient and records its conditions
challengeSolves a disguised quadratic in a logarithmic or exponential quantity and rejects one branch
examCombines an exact solution choice, coefficient table and restriction choice

Questions about this topic

What is the exponential form of a logarithmic equation?+

The argument equals the base raised to the logarithm's value, provided the base is positive and a is not 1 and the argument u is positive

Why must the coefficient m be non-zero?+

The rearranged solution divides by m. Division by zero is not defined, so m cannot be zero

Why can a quadratic logarithmic equation have an inadmissible branch?+

Algebra may produce a candidate that makes an original logarithm argument non-positive. Each candidate must be checked in the original equation

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 4: Logarithmic functions

In this chapter

Connect exponential and logarithmic formsApply logarithm laws and inverse identitiesEvaluate logarithms and antilogarithmsExpress one logarithmic quantity in terms of anotherUse a formula-bearing logarithmic scaleSolve exponential equations and inequalities exactlySolve exponential and logarithmic equationsSolve literal logarithmic and indicial equationsAnalyse exponential and logarithmic graphsConstruct exponential and logarithmic inverse pairsInterpret and use an exponential modelInterpret and use a logarithmic modelDetermine and use a logarithmic model