# Thread: Solving for x in an exponential function

1. ## Solving for x in an exponential function

I've been struggling on figuring out these kinds of problems when I need to set f(x) equal to zero and solve for x.

$f(x)={12(1.5^x)}+{12(0.5^x)}$

Here's my thought process:

1) $0={12(1.5^x)}+{12(0.5^x)}$

2) $\frac{0}{12}=\frac{12(1.5^x)}{12}+\frac{12(0.5^x)} {12}$

3) ${0}={(1.5^x)}+(0.5^x)}$

4) ${0}={(1.5^x)}+(0.5^x)}$; here's where my intuition breaks down. I want to take the log of both sides but I can't due to the zero! Am I going about this kind of problem wrong from the beginning?

2. ## Re: Solving for x in an exponential function

$1.5^x + 0.5^x = 0$ has no solution ... why?

3. ## Re: Solving for x in an exponential function

No logs of 0. Oh, so this is why there's always a vertical asymptote at x=0 for logarithmic functions.

4. ## Re: Solving for x in an exponential function

Originally Posted by AZach
No logs of 0. Oh, so this is why there's always a vertical asymptote at x=0 for logarithmic functions.
no ...

$1.5^x > 0$

$0.5^x > 0$

how can you add two positives and get 0 ?