lg equation

**paulaa** · January 10th 2011, 12:04 PM

does someone know how to salve this ?

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

**pickslides** · January 10th 2011, 12:25 PM

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

**Also sprach Zarathustra** · January 10th 2011, 12:36 PM

Originally Posted by paulaa

does someone know how to salve this ?

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

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

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

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

.
.
.

**CaptainBlack** · January 10th 2011, 01:10 PM

Originally Posted by Also sprach Zarathustra

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

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

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

.
.
.

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

CB

Thread: lg equation