Exponential using two different bases.
8x-1 = 9x+3
Solve for x.
I'm guessing since they're both equal, if I multiply them it will just be the same as squaring one side.
We have (8x-1)(9x+3)=72x^2+2x-3
ANNNNND I don't know what to do next. I tried equating x2+2x-3 = 2 since 72x^2+2x-3 = a2 where a is the base. Solving for x2+2x-5=0, I got:
How do I solve this? I need to find the base.
Re: Exponential using two different bases.
Recall that you can calculate the log in any base by log_b(x) = log_c(x)/log_c(b) where log_c(x) is the logarithm in base c.
This means that you can calculate log_9(x) in terms of log_8(x) and vice versa.