SOLUTION: Can you please help me with this question. The problem is 5^x-2=4^x-1. The answer I came up with is log4 but that doesnt seem right can you help? Thanks

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Can you please help me with this question. The problem is 5^x-2=4^x-1. The answer I came up with is log4 but that doesnt seem right can you help? Thanks       Log On


   

Question 108433: Can you please help me with this question.
The problem is 5^x-2=4^x-1. The answer I came up with is log4 but that doesnt seem right can you help? Thanks

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
If it seems right, it should solve the equation.
Plug the value in the equation and check the result.
5%5E%28x-2%29=4%5E%28x-1%29
5%5E%28log%284%29-2%29=4%5E%28log%284%29-1%29
5%5E%28.6021-2%29=4%5E%28.6021-1%29
5%5E%28-1.3979%29=4%5E%28-0.3979%29
.10542=.57602
Not a true statement. Not a solution.
Let's look at the problem.
5%5E%28x-2%29=4%5E%28x-1%29
Let's use some variables now.
Let A=5%5E%28x-2%29 and let B=4%5E%28x-1%29
5%5E%28x-2%29=4%5E%28x-1%29
A=B
log%285%2CA%29=log%285%2CB%29
Also, you know from your log rules that
log%285%2CA%29=log%284%2CB%29%2Flog%284%2C5%29
If you put that all together you get,
log%285%2CA%29=log%285%2CB%29
log%285%2CA%29=log%284%2CB%29%2Flog%284%2C5%29
log%284%2C5%29%2Alog%285%2CA%29=log%284%2CB%29
Now back to A and B.
A=5%5E%28x-2%29
log%285%2CA%29=x-2
B=4%5E%28x-1%29
log%284%2CB%29=x-1
Substituting,
log%284%2C5%29%2Alog%285%2CA%29=log%284%2CB%29
log%284%2C5%29%2A%28x-2%29=%28x-1%29
Let's define a constant, c,
c=log%284%2C5%29=1.161
c%28x-2%29=%28x-1%29
cx-2c=x-1
cx-x=2c-1
x%28c-1%29=2c-1
x=%282c-1%29%2F%28c-1%29
Now substituting for the real values,
x=%282%281.161%29-1%29%2F%281.161-1%29
x=8.213
Check your answer.
5%5E%28x-2%29=4%5E%28x-1%29
5%5E%288.213-2%29=4%5E%288.213-1%29
5%5E%286.213%29=4%5E%287.213%29
22014=22011
Close enough from roundoff error.
x=8.213