# `62=116log(a+40)-176`Find `a`.

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To solve this problem, you must first determine the base of the logarithm. If you are a math major in college, then `log = ln`. However, I will proceed as if you mean `log_10` (I apologize if you mean `ln`!) because that base is more applicable to members of the site.

So, let's solve the following equation:

`62 = 116log(a+40) - 176`

We'll start like we would for any other equation and isolate the logarithm first. First, let's add 176 to both sides:

`238 = 116log(a+40)`

Now, let's divide by 116:

`238/116 = log(a+40)`

In order to get rid of the logarithm, we must allow both sides to be exponents of 10:

`10^(238/116) = 10^(log(a+40))`

Because log and 10^ are inverse functions, the 10^ and log functions cancel out on the right:

`10^(238/116) = a + 40`

Finally, we can subtract 40 from both sides to find `a`:

`10^(238/116) - 40 = a`

Plugging this into the calculator should give you the following result:

`a ~~ 72.65`

I hope this helps!

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