



 
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
 


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