I'm not sure how to start this problem I'm able to get it to (4x-3)log(4)=(9x-6)log(3). Not sure how to continue.

Question

Original question 4^(4x-3)=3^(9x-6)

Answer 1

the logs are just (messy) numbers that you can distribute, add, subtract...

[4 log(4)] x - 3 log(4) =
... [9 log(3)] x - 6 log(3)

grind away...

Answer 2

After a while, it's just plain algebra.

Answer 3

4 ^ ( 4 x - 3 ) = 3 ^ ( 9 x - 6 )

( 2 ^ 2 ) ^ ( 4 x - 3 ) = 3 ^ ( 9 x - 6 )

Take the natural logarithm of both sides and use the identity:
log ( a ^ b ) = b * log a

log 2 ^ 2 = 2 * log ( 2 )

log [ ( 2 ^ 2 ) ^ ( 4 x - 3 ) ] =

log ( 2 ^ 2 ) * ( 4 x - 3 ) =

2 * log ( 2 ) * ( 4 x - 3 )

log [ 3 ^ ( 9 x - 6 ) ]=

log ( 3 ) * ( 9 x - 6 )

So 4 ^ ( 4 x - 3 ) = 3 ^ ( 9 x - 6 ) mean:

2 * log ( 2 ) * ( 4 x - 3 ) = log ( 3 ) * ( 9 x - 6 )

2 * log ( 2 ) * 4 x - 2 * log ( 2 ) * 3 = log ( 3 ) * 9 x - log ( 3 ) * 6

8 log ( 2 ) * x - 6 log ( 2 ) = 9 log ( 3 ) * x - 6 log ( 3 ) Add 6 log ( 2 ) to both sides

8 log ( 2 ) * x - 6 log ( 2 ) + 6 log ( 2 ) = 9 log ( 3 ) * x - 6 log ( 3 ) + 6 log ( 2 )

8 log ( 2 ) * x = 9 log ( 3 ) x - 6 log ( 3 ) + 6 log ( 2 ) Subtract 9 log ( 3 ) * x to both sides

8 log ( 2 ) * x - 9 log ( 3 ) * x = 9 log ( 3 ) x - 6 log ( 3 ) + 6 log ( 2 ) - 9 log ( 3 ) * x

8 log ( 2 ) * x - 9 log ( 3 ) * x = - 6 log ( 3 ) + 6 log ( 2 )

x * [ 8 log ( 2 ) - 9 log ( 3 ) ] = 6 log ( 2 ) - 6 log ( 3 ) Divide both sides by [ 8 log ( 2 ) - 9 log ( 3 ) ]

x = 6 [ log ( 2 ) - log ( 3 ) ] / [ 8 log ( 2 ) - 9 log ( 3 ) ]

x = 0.56025