"Solve for x:"

2 X 3^2x+3 = 3 X 2^3x+4

(X is multiply)

feel silly for posting this, is a simple log question i think, but its got me stumped.

thanks

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- Jul 15th 2010, 10:52 AMporge111A rather basic log question I believe
"Solve for x:"

2 X 3^2x+3 = 3 X 2^3x+4

(X is multiply)

feel silly for posting this, is a simple log question i think, but its got me stumped.

thanks - Jul 15th 2010, 11:09 AMAckbeet
Just to clarify, you want to solve Is that correct? If so, what steps have you taken?

- Jul 15th 2010, 11:35 AMbigwave
it could be also:

in which case - Jul 15th 2010, 11:55 AMporge111
sorry yeah the above is the way it's laid out - how did you get -1 ?

- Jul 15th 2010, 12:09 PMbigwavewell several ways to solve this..
well several ways to solve this..

but you can divide both sides by and to elimate the coeficients,

then

after calculation we are left with

- Jul 15th 2010, 12:37 PMporge111
ahhh it's so simple, i'm so stupid. haha thanks, i just didn't start with the basic, "dividng by 2" part. oops.

- Jul 15th 2010, 10:51 PMZharif93
2(3)^2x+3 = 3(2)^3x+4

log 2 × log 3^(2x+3) = log 3 × log 2^(3x+4)

(2x+3) log 3 = log 3

(3x+4) log 2 = log 2

2x log 3 + 3 log = log 3

2x log 3 = log 3 - 3 log 3

x(2 log 3) = -2 log 3

x = -2 log 3

/2 log 3

x = -1