# Exponential/Logarithmic system of equations

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!

• When you eliminate $c$ in equations 2 and 3, you obtain $2^{ab}=256$ which is true for $ab=8$, which is your first equation – imranfat Nov 8 '15 at 3:59
• Awesome. So from there, I get $a = 8$, $b = 1$, and $c = 256$. What's more, I can also get $a = 1$, $b = 8$, $c = 2$. Make this an answer and I'll mark you correct. – ollien Nov 8 '15 at 4:05
• Double checking my work with Wolfram, it looks like there's a couple other solutions I'm missing actually. Specifically, $a = 2$, $b = 4$, $c =4$, and $a= 2$, $b = 4$, $c = 16$. How can I get to these solutions? Anything I seem to try proves futile. – ollien Nov 8 '15 at 4:08
• You can take any values you want for $a$ and $b$ as long as $ab=8$. Then let $c=2^a$ and you'll have a solution for the system. The reason that you couldn't get any further is because $(\log_2 c)(\log_c 256)=8$ is true for all positive values of $c$! – Dylan Nov 8 '15 at 4:17
• I think my answer pretty much coincides with Claude's...you can mark his correct... – imranfat Nov 8 '15 at 4:43

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