How do you solve: 7^2=12^x

x = 1.566

Step-by-step explanation:

7^2=12^x

49 = 12^x

Exponential: b^x = y

Logarithmic: logb(y) = x.

logb(y) = x

log12(49) = x

Plug this in your calculator (on a TI-84, it's MATH -> A)

x = 1.566

x ≈ 1.57

Step-by-step explanation:

First, note that 7^2 equals 7 * 7 = 49. Now, we have:

49 =

In order to solve for x, we need to use logarithm, which basically "gets rid of" the exponent. What we need to do is take the of each side. Notice that the base of the logarithm is 12. This is because we want to cancel out the exponent base of 12 on the right side:

So, x ≈ 1.57.

Hope this helps!

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