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Solving Exponential Equations Using Logarithms

In order to solve exponential equations algebraically you need to use logaritms.

Exponential equations with base 10

The following method is used to solve exponential equations where the powers have a base-10, like for example 210x=62.2\cdot10^x = 62.

Solve for the power with the unknown variable so that it is by itself on either the right or left hand side.
210x=622 \cdot 10^x = 62
10x=3110^x = 31
By using log on the both sides and simplifying, you can separate out the variable so that it is along on one side of the equation.
10x=3110^x = 31
lg(10x)=lg(31)\lg \left( 10^x \right) = \lg(31)
lg(10a)=a \lg\left(10^a\right)=a
x=lg(31)x = \lg(31)

This is the exact solution for the equation. However, if you are not asked for the exact answer, you can get an approximate one by entering lg(31)\lg(31) into your calculator: lg(31)1.49. \lg(31) \approx 1.49.

General exponential equations

Exponential equations with arbitrary bases, like 2x1=982^x-1=98, for example, can be solved using the power rule of logarithms.

Solve for the term that includes the power with the unknown variable so that it stands alone on one side.
2x1=982^x-1= 98
2x=992^x = 99

Take the logaritm of the left and right hand sides.

2x=992^x = 99
lg(2x)=lg(99)\lg \left( 2^x \right) = \lg (99)
Pull down the exponent and place it in front of the logarithm.
lg(2x)=lg(99)\lg \left( 2^x \right) = \lg (99)
xlg(2)=lg(99)x \cdot \lg (2) = \lg (99)
Solve for the variable by dividing by the logarithm.
xlg(2)=lg(99)x \cdot \lg (2) = \lg (99)
x=lg(99)lg(2)x = \dfrac{\lg (99)}{\lg (2)}

This is the answer in its exact form, and if you enter it into your calculator you will get the approximate answer x6.63.x \approx 6.63.