Hi Todd,

Let me show you a similar question: 3(2^x+4)=5e^x

Since there is no way we can change the bases of the exponential we need to take the logarithm of both sides of the equations (For a review: the relationship between logarithms and exponential functions.) I will use the natural logarithm where ln is the same thing as log_e because there is an e in my equation.

ln[3(2^x+4)]=ln(5e^x)

Using the law of logarithms where log ab=log a + log b, I simplify the equation:

ln 3 + ln (2^x+4) = ln 5 + ln(e^x)

Using the law of logarithms where log a^b = b log a, I will simplify further:

ln 3 + (x+4)(ln 2) = ln 5 + x (ln e)

I will distribute ln 2 through x+4 and simplify ln e since it equals 1:

ln 3 + x ln2 + 4 ln2 = ln 5 + x

Now I will collect the terms with variables to one side of the equation and those without to the other side:

x ln2 - x = ln 5 -ln 3 - 4 ln2

I need to get x by itself so I will factor it out:

x (ln2 - 1) = ln 5 -ln 3 - 4 ln2

Now I will divide both sides by ln2 -1:

x = (ln 5 -ln 3 - 4 ln2)/(ln2 -1)

You can keep this value of x for an exact value or enter it into your calculator to find the approximate value.

Hope this helps,

Janice