# Math Help - Logarithmic equations

1. ## Logarithmic equations

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

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

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

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

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