Question

Solve 12^x-2=20 round to the nearest ten-thousandth? help, please!

Answer 1

The answer to the equation: 12^x-2=20 round to the nearest ten-thousandth is:

`x=1.2440`

Explanation:

We need to find x of the equation (1)

`12^x-2=20`

Now, the number two goes to the other side of the equation to add

`12^x=20+2\\12^x=22`

We need to multiply by logarithm on both sides of the equation

`Log(12^x)=Log(22)`

Using the properties of the logarithms, we know:

`x*Log(12)=Log(22)\\x=(Log22)/(Log12)`

`x=1.2439`

The nearest ten-thousandth is:

`x=1.2440`

Answer 2

12^x - 2 = 20
12^x = 22
x log 12 = log 22
x = (log 22)/(log 12)

The rest is a calculator exercise. Use any base log you like, as long as both are the same base. (The right side is really the base-12 logarithm of 22, but nobody has a base-12-log key on their calculator.)