1. ## 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

2. ## 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 $\displaystyle log_4(x)(log_8(x))= log_16(x)$.

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

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

If we let $\displaystyle y= log_2(x)$ that is the same as $\displaystyle \frac{y}{2}\frac{y}{3}= \frac{y^2}{6}= \frac{y}{4}$.
Can you solve that?

3. ## Re: log

Do you mean:

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

If so, try the identity:

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

4. ## Re: log

Originally Posted by MarkFL2
Do you mean:

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

If so, try the identity:

$\displaystyle \log_{b^n}(a)= \frac{{\log_b(a)}}{n}$
nvm did not see halloy..

5. ## Re: log

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.