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Det här är en översatt version av sidan Lösa exponentialekvationer med logaritmer *Method*. Översättningen är till 100% färdig och uppdaterad.

# Solving Exponential Equations Using Logarithms

In order to solveexponential 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 of 10 like for example $2\cdot10^x = 62.$

Solve for the power with the unknown variable so that it is by itself on either the right or left hand side.
$2 \cdot 10^x = 62$
$10^x = 31$
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.
$10^x = 31$
$\lg \left( 10^x \right) = \lg(31)$
$\lg\left(10^a\right)=a$
$x = \lg(31)$

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

## General exponential equations

Exponential equations with arbitrary bases, like for $2^x-1=98$ for example, can be solved using the logarithmic power laws.

Solve for the power with unkown variable so that it stands alone on one side.
$2^x-1= 98$
$2^x = 99$

Take the logaritm of the left and right hand sides.

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

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