Exponential/Logarithmic system of equations

Question

I'm given the following equations of which I need to find the solutions for algebraically.

$ab = 8$

$2^a=c$

$c^b = 256$

My first thought was to use logarithms, but I got a bit lost in doing so

$\log_2 2^a = \log_2 c$

$a = \log_2 c$

$\log_c c^b = \log_c 256$

$b = \log_c 256$

And plugging those into the first equation I get

$(\log_2 c)(\log_c 256) = 8$

However from here, I'm at a loss on what to do. Can someone help me out here? Thanks!

Answer 1

Considering the second equation $2^a=c$, you have $a=\log_2(c)$. From the third equation $c^b=256$, $b\log_2(c)=\log_2(256)=8$. To summarize $$a=\log_2(c)$$ $$b\log_2(c)=8$$ Making the product $$a b\log_2(c)=8\log_2(c)$$ But in the first equation $ab=8$; so the last expression is satisfied for any value of $c$ as long as $c\ne 1$ (since $\log_2(1)=0$ and you cannot divide by $0$) and any couple $(a,b)$ as long as $ab=8$.

Answer 2

Hint: Given that $ab=8$ what does that say about the possible values for $a$ and $b$?