Solving equations where the variable is in an exponent usually involves use of logarithms. You start by isolating the base and its exponent. So we'll start by adding four to each side:
and then dividing both sides by 3:
Now the base and its exponent are isolated. Next we findthe logarithm of each side. You can use any base of logarithm but it is to your advantage to use a base that matches the base of the exponent. So we'll use base e (aka ln) logarithms:
Now we use a property of logarithms, , which allows us to move the exponent of an argument out in front. It is this very property which is the reason for using logarithms on these problems. It gives us a valid way to get the variable out of the exponent!
Using this property on our equation we get:
By definition, ln(e) = 1. (This is why base e logarithms were the best choice). So this simplifies to:
The only thing left is to solve for x. Multiply (or divide) both sides by -1:
This is an exact solution for x. If you want a decimal approximation, then get out your calculator and...
Calculate 13/3
Find the ln of the result form step 1
Change the sign of the result from step 2.
5log(x-2)=11
As you might guess, we use a similar process on this equation but we use exponentiation to get the variable out of the argument of the logarithm. First we isolate the logarithm by dividing both side by 5:
The logarithm is isolated. Now we rewrite the equation in exponential form:
Now that the x is out where we can "get at it", we can solve for it. (While it remained in the argument of a logarithm, we would not be able to solve for x.) Adding 2 to each side we get:
This is an exact expression for the solution. If you want a decimal approximation, then get out your calculator and have it calculate the right side.
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