# log

• Nov 15th 2012, 08:21 AM
Petrus
log
ima solve al none negative solution för this equation ehmm idk how to write it but its ^(4) *logx*^(8)*log x = ^(16)*log x idk how to solve this:(
• Nov 15th 2012, 12:21 PM
HallsofIvy
Re: log
I'm not sure what the "^4" means but I will guess that it is the base of the logarithm. I am used to that being written as a subscript, after. If that is correct then the equation is $log_4(x)(log_8(x))= log_16(x)$.

Now, " $y= log_a(x)$" is the same as " $a= e^y$". Do you notice that all of those bases are powers of 2? $4= 2^2$, $8= 2^3$ and $16= 2^4$. So $y= log_4(x)$ is the same as $x= 4^y= (2^2)^y= 2^{2y}$ and so $log_2(x)= 2y$, thus $y= log_{4}(2)= \frac{log_2(x)}{2}$. $y= log_8(x)$, so $x= 8^y= (2^3)^y= 2^{3y}$, thus $3y= log_2(x)$ and $y= log_8(x)= \frac{log_2(x)}{3}$. Similarly, $log_{16}(x)= \frac{log_2(x)}{4}$ so the equation can be written

$\frac{log_2(x)}{2}\frac{ln_2(x)}{3}= \frac{ln_2(x)}{16}$

If we let $y= log_2(x)$ that is the same as $\frac{y}{2}\frac{y}{3}= \frac{y^2}{6}= \frac{y}{4}$.
Can you solve that?
• Nov 15th 2012, 12:25 PM
MarkFL
Re: log
Do you mean:

$\log_4(x)\cdot\log_8(x)=\log_{16}(x)$ ?

If so, try the identity:

$\log_{b^n}(a)= \frac{{\log_b(a)}}{n}$
• Nov 15th 2012, 12:27 PM
Petrus
Re: log
Quote:

Originally Posted by MarkFL2
Do you mean:

$\log_4(x)\cdot\log_8(x)=\log_{16}(x)$ ?

If so, try the identity:

$\log_{b^n}(a)= \frac{{\log_b(a)}}{n}$

nvm did not see halloy..
• Nov 15th 2012, 12:45 PM
skeeter
Re: log
Quote:

Originally Posted by Petrus
ima solve al none negative solution för this equation ehmm idk how to write it but its ^(4) *logx*^(8)*log x = ^(16)*log x idk how to solve this

please do not use "text talk" abbreviations ... it makes the post confusing and difficult to read.