quick help on solving for 'x' with log

[laura.edan]

hey i'm stuck on this one question...
Log[sub:38s61o4l]7[/sub:38s61o4l]x+log[sub:38s61o4l]7[/sub:38s61o4l](x-1)=log[sub:38s61o4l]7[/sub:38s61o4l]2x

what i got so far is
log[sub:38s61o4l]7[/sub:38s61o4l]x + log[sub:38s61o4l]7[/sub:38s61o4l](x-1) - log[sub:38s61o4l]7[/sub:38s61o4l]2x = 0

please help!
thanks

 

[Loren]

log (AB) = log A + log B
and
log (A/B) = log A - log B

Use these backwards for your problem???

 

[laura.edan]

so would that become...
log[sub:2bzw1z4u]7[/sub:2bzw1z4u] ((x)(x-1))/ 2x = 0

 

[skeeter]

log[sub:e46nx4em]7[/sub:e46nx4em]x + log[sub:e46nx4em]7[/sub:e46nx4em](x-1) = log[sub:e46nx4em]7[/sub:e46nx4em](2x)

log[sub:e46nx4em]7[/sub:e46nx4em][x(x-1)] = log[sub:e46nx4em]7[/sub:e46nx4em](2x)

if log(a) = log(b), then a = b

so ...

x(x-1) = 2x

now solve the quadratic ... be sure to check your solutions in the original equation for domain issues.