1. ## Logarithmic equations

Two similar logarithmic equations:
1) $\displaystyle x^{lgx}=1000x^2$
2) $\displaystyle x^{\frac{lgx+7}{4}}=10^{lgx+1}$

I tried solving the first one:
$\displaystyle x>0;$
I tried to write the first exponent in the base x, so that I could apply the rule $\displaystyle a^{loga(b)}=b$:
$\displaystyle x^{\frac{logx(x)}{logx(10)}}=1000x^2$
But it didn't work, because I got in the exponent $\displaystyle \frac{1}{logx(10)}$
How do I solve these?

2. Originally Posted by Evaldas
Two similar logarithmic equations:
1) $\displaystyle x^{lgx}=1000x^2$
Take $\displaystyle lgx$ to be $\displaystyle \log_{10}(x)$
Then the equation becomes $\displaystyle \left(\log(x)\right)^2-2\log(x)-3=0$.
Let $\displaystyle u=\log(x).$
Can you solve $\displaystyle u^2-2u-3=0~?$

3. Originally Posted by Plato
Then the equation becomes $\displaystyle \left(\log(x)\right)^2-2\log(x)-3=0$.
How do you get this?

4. Originally Posted by Evaldas
How do you get this?
$\displaystyle \log\left(x^{\log(x)}\right)=\log\left(1000x^2\rig ht)$.
Apply the rules of logarithms.