SOLUTION: solve the equation to the nearest thousandth: 2^x+1=12

Question 394362: solve the equation to the nearest thousandth:
2^x+1=12
Answer by jsmallt9(3759) (Show Source): You can put this solution on YOUR website!
I'm guessing that the equation is:

and not

If this is correct, then please use parentheses on multiple term exponents. For example:
2^(x+1) = 12
Tutors are more likely to help if the problem is stated clearly.

Solving equations where the variable is in an exponent usually involves logarithms. The base of the logarithm used does not make a significant difference. But if you want a decimal approximation (and you problem asks for a decimal), then you should choose a base of logarithm that your calculator "knows", like base 10 or base e (aka ln).

I'm going to use base e logarithms. (If you choose base 10 instead the answer works out the same!)

On the left side we can use a property of logarithms, , to move the exponent in the argument out in front. It is this very property that is the reason we use logarithms on problems like this. By moving the exponent, where the variable is, out in front the variable can now be solved for. Using this property of our equation we get:

Now we can solve for x. Dividing by ln(2) we get:

Subtracting 1 we get:

This is an exact expression for the answer. For the decimal approximation we go to our calculators to find the two logarithms:

x = 3.5849625007211562 - 1
which simplifies to:
x = 2.5849625007211562
Rounded to the nearest thousandth we get:
x = 2.585
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