Check out some similar questions!

BigTreeWoody · Jun 3, 2009, 07:22 PM

1/2log[base64] - log[basex]=log6

1/2log[base64] - log[basex]=log4

Perito · Jun 3, 2009, 07:34 PM

$\frac 12 log_{64}(?) - log_x(?)=log_{10}(6)$

$\frac 12 log_{64}(?) - log_x(?)=log_{10}(4)$

I think you're missing something in your equations. At least you are if I understood what you wrote.

BigTreeWoody · Jun 3, 2009, 07:37 PM

Woops! 64 and x are not bases 1/2log64-logx=log 6
1/2log64-logx=log4

Perito · Jun 3, 2009, 07:42 PM

Originally Posted by BigTreeWoody

Woops! 64 and x are not bases 1/2log64-logx=log 6
1/2log64-logx=log4

$\frac 12 log(64) - log(x)=log(6)$

$\frac 12 log(64) - log(x)=log(4)$

1. Recognize that $a\,log(b)=log(b^a)$ .

2. Recognize that $a^{\frac 12} = \sqrt{a\,}$

3. Isolate the log(x) term on one side of the equation.

4. Take the antilog $antilog(log(x))=10^{log(x)}=x$ and solve for x.