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Use logarithms to solve exponential equations

Logarithms, as inverses of exponential functions, enable solving equations such as bx=kb^x=k and Abmx+n=CA b^{mx+n}=C by isolating the exponential expression, taking a logarithm of both sides, and applying logarithm properties. Bases must be positive and not equal to 1, logarithm arguments must be positive, and solutions may be exact or approximated with common or natural logarithms; logarithms do not distribute over sums, and systems, complex-valued cases, and more advanced nonlinear forms are outside this scope.

Detailed Explanation: Use logarithms to solve exponential equations

To solve an exponential equation, first isolate the exponential expression. Then take a logarithm of both sides. Since logarithms and exponentials are inverse operations, the exponent can be brought down using the power property:

log⁡(ar)=rlog⁡(a).\log(a^r)=r\log(a).

Example: Solve

5⋅32x−1=40.5\cdot 3^{2x-1}=40.

1. Isolate the exponential expression.
Divide both sides by 55:

32x−1=8.3^{2x-1}=8.

2. Take a logarithm of both sides.
You may use common logarithms or natural logarithms:

log⁡(32x−1)=log⁡(8).\log\left(3^{2x-1}\right)=\log(8).

3. Bring the exponent down.

(2x−1)log⁡(3)=log⁡(8).(2x-1)\log(3)=\log(8).

4. Solve for xx.
Divide by log⁡(3)\log(3):

2x−1=log⁡(8)log⁡(3).2x-1=\frac{\log(8)}{\log(3)}.

Add 11 and divide by 22:

x=1+log⁡(8)log⁡(3)2.x=\frac{1+\frac{\log(8)}{\log(3)}}{2}.

This is an exact logarithmic answer. Using a calculator,

x≈1+1.89282≈1.446.x\approx \frac{1+1.8928}{2}\approx 1.446.

So the solution is

x≈1.446.\boxed{x\approx 1.446}.

Remember that logarithm arguments must be positive, so you can only take log⁡\log of a positive quantity. Also, logarithms do not distribute over sums: log⁡(a+b)≠log⁡(a)+log⁡(b)\log(a+b)\ne\log(a)+\log(b).

Learn by doing: Use logarithms to solve exponential equations

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