[mickey97]

For example, solving the equasion y = 2^(x-4) for x yields x = y^(1/2) + 4. But I can't seem to do the same for y = 6^(x+1) - 6

[ebaines]

Your example is wrong. Consider if x = 5; then y = 2^(x-4)= 2^1 = 2. But your "solution" says x = y^(1/2) + 4 = sqrt(2) + 4, which is certainly not equal to 5. Your error here is in thinking that the square root of 2^(x-4) is x-4.

Why your aversion to using logs?

[harum]

I doubt you can avoid logarithms in your problem. A neat trick might help you calculate logarithms faster, if this is why you choose to avoid them:

(log(base b) a) = (log(base c) a) / (log(base c) b);

For example, (log(base 6) 10) = ln(10) / ln(6), or = lg(10)/lg(6).
"ln" is log(base e), "lg" is log(base 10).

[galactus]

Factor:



Now, continue and use logs.