1. ## lg equation

does someone know how to salve this ?

x^(1+lg(x))=(10x)^-2

2. Using guess and check on powers of 10 you'll find x = 1/10 will do the job.

3. Originally Posted by paulaa
does someone know how to salve this ?

x^(1+lg(x))=(10x)^-2
$x^{(1+log(x))}=(10x)^{-2}$

$logx^{(1+log(x))}=log(10x)^{-2}$

$logx(1+logx)=-2(1+logx)$

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4. Originally Posted by Also sprach Zarathustra
$x^{(1+log(x))}=(10x)^{-2}$

$logx^{(1+log(x))}=log(10x)^{-2}$

$logx(1+logx)=-2(1+logx)$

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$\text{lg}$ is ambiguous it can denote a base 10 or base 2 logarithm

CB