# Thread: Simultaneous Equations Logarithms Problem

1. ## Simultaneous Equations Logarithms Problem

the base is inside the ()

y=2log(3)x

y+1=log(3)9x

what i did:

y=2log(3)x
y=log(3)9x-1

2log(3)x = log(3)9x-1
2log(3)x = 2log(3)3X - log(3)3
2log(3)x - 2log(3)3x = - log(3)3
x(2log(3)1 - 2log(3)3) = -log(3)3
x = -log(3)3 / (2log(3)1 - 2log(3)3)
x = -1 / (0-2)
x = 1/2

could you tell me hwere i went wrong please?

2. Originally Posted by juliak
the base is inside the ()

y=2log(3)x

y+1=log(3)9x

what i did:

y=2log(3)x
y=log(3)9x-1

2log(3)x = log(3)9x-1
2log(3)x = 2log(3)3X - log(3)3
2log(3)x - 2log(3)3x = - log(3)3
x(2log(3)1 - 2log(3)3) = -log(3)3
x = -log(3)3 / (2log(3)1 - 2log(3)3)
x = -1 / (0-2)
x = 1/2

could you tell me hwere i went wrong please?
Hi juliak,

Here we go....

$\displaystyle 2\log _3 (x)=\log_3 (9x)-1$

$\displaystyle 2\log_3 (x)=\log_3(9)+\log_3(x)-1$

$\displaystyle 2\log_3(x)=2+\log_3(x)-1$

$\displaystyle 2\log_3(x)=\log_3(x)+1$

$\displaystyle \log_3(x)=1$

$\displaystyle x=3^1$

$\displaystyle x=3$