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Solve exponential equations using logarithms

Solving exponential equations with logarithms involves using the inverse relationship between exponential and logarithmic functions to isolate an exponent, including equations such as akx+c=ba^{kx+c}=b and Aakx=CA a^{kx}=C, while applying logarithm properties correctly and respecting positive bases and arguments. The solution may be expressed exactly with logarithms or approximated numerically, and must be checked against the original equation; complex solutions and advanced generalized equations are not included.

Detailed Explanation: Solve exponential equations using logarithms

To solve an exponential equation, use a logarithm to bring the exponent down where you can isolate it.

Consider the example

32x−1=20.3^{2x-1}=20.

  1. Take the logarithm of both sides.
    We can use either common logarithms or natural logarithms:

log⁡(32x−1)=log⁡(20). \log\left(3^{2x-1}\right)=\log(20).

  1. Use the power property of logarithms, log⁡(ar)=rlog⁡(a)\log(a^r)=r\log(a):

(2x−1)log⁡(3)=log⁡(20). (2x-1)\log(3)=\log(20).

  1. Divide by log⁡(3)\log(3) to isolate the exponent expression:

2x−1=log⁡(20)log⁡(3). 2x-1=\frac{\log(20)}{\log(3)}.

  1. Solve for xx:

2x=1+log⁡(20)log⁡(3) 2x=1+\frac{\log(20)}{\log(3)} x=1+log⁡(20)log⁡(3)2. \boxed{x=\frac{1+\frac{\log(20)}{\log(3)}}{2}}.

Using a calculator,

x≈1.863. \boxed{x\approx 1.863}.

  1. Check the solution in the original equation:

32(1.863)−1≈32.726≈20. 3^{2(1.863)-1}\approx 3^{2.726}\approx 20.

The solution is reasonable because it makes the original equation true.

For an equation such as akx+c=ba^{kx+c}=b, the general process is

akx+c=b⟹log⁡(akx+c)=log⁡(b)a^{kx+c}=b \quad\Longrightarrow\quad \log(a^{kx+c})=\log(b) (kx+c)log⁡(a)=log⁡(b),(kx+c)\log(a)=\log(b),

then solve the resulting linear equation for xx. The base must satisfy a>0a>0 and a≠1a\ne1, and the logarithm’s argument must be positive.

Learn by doing: Solve exponential equations using logarithms

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