# Thread: [SOLVED] Proving log problem

1. ## [SOLVED] Proving log problem

Show that:

$\displaystyle \log_{x^2}(y)$ = $\displaystyle \log_x\sqrt{y} \$

$\displaystyle x > 0$ and $\displaystyle y > 1$

2. $\displaystyle \log_{x^2}y=\frac{\log_xy}{\log_xx^2}=\frac{\log_x y}{2}=\frac{1}{2}\log_xy=\log_x(y)^{\frac{1}{2}}=\ log_x\sqrt{y}$

3. Originally Posted by MarcoMP
Show that:

$\displaystyle \log_{x^2}(y)$ = $\displaystyle \log_x\sqrt{y} \$

$\displaystyle x > 0$ and $\displaystyle y > 1$

Use the base change formula:

$\displaystyle \log_a(x)=\dfrac{\ln(x)}{\ln(a)}$

\displaystyle \begin{aligned}\log_{x^2}(y)&= \\ &= \dfrac{\ln(y)}{\ln(x^2)} \\ &=\dfrac{\ln(y)}{2\ln(x)} \\ &=\dfrac{\frac12 \ln(y)}{\ln(x)} = \dfrac{\ln(\sqrt{y})}{\ln(x)} = \log_x(\sqrt{y})\end{aligned}

4. Could you please explain why both bases are $\displaystyle x$ when you put in fraction form?

earboth could you do that using only logs laws seeing as I dont understand the $\displaystyle In$ yet

5. Originally Posted by MarcoMP
Could you please explain why both bases are $\displaystyle x$ when you put in fraction form?

earboth could you do that using only logs laws seeing as I dont understand the $\displaystyle In$ yet
Have a look at red-dog's solution: He only uses $\displaystyle \log_x$ and he provides you with a much better and elegant solution than I did.

6. $\displaystyle \log_{x^2}y=\frac{\log_{10}{y}}{\log_{10}{x^2}}=\f rac{\log_{10}y}{2log_{10}x}=\frac{1}{2}\log_xy=\lo g_x(y)^{\frac{1}{2}}=\log_x\sqrt{y}$

Is that correct?

7. Originally Posted by MarcoMP
$\displaystyle \log_{x^2}y=\frac{\log_{10}{y}}{\log_{10}{x^2}}=\f rac{\log_{10}y}{2log_{10}x}=\frac{1}{2}\log_xy=\lo g_x(y)^{\frac{1}{2}}=\log_x\sqrt{y}$

Is that correct?
Perfect!

8. Thanks very much! Now just got to get $\displaystyle x$ in:

$\displaystyle e^{x^2} = \frac{1}{\sqrt {3^{x^5}}}$ I'm quite lost i must say

9. Originally Posted by MarcoMP
Thanks very much! Now just got to get $\displaystyle x$ in:

$\displaystyle e^{x^2} = \frac{1}{\sqrt {3^{x^5}}}$ I'm quite lost i must say
1. Please do yourself and do us a favour and start a new thread if you have a new question.

2. This one looks very ugly to me. Are you sure you typed it correctly here? (I'm only asking because the key "e" is exactly under the key"3" and maybe you slipped a little bit?) If the e is meant to be a 3 then your question becomes:

$\displaystyle 3^{x^2} = \frac{1}{\sqrt {3^{x^5}}} ~\implies~ 3^{x^2} \cdot \sqrt {3^{x^5}}= 1~\implies~3^{x^2+\frac52 x^2} = 1=3^0~\implies~\frac72x^2=0~\implies~x=0$