SOLUTION: Solve exactly for x if possible. If not, find an approximate solution (two decimal places). 3*2^(5x+3)=7*6^x Any help would be greatly appreciated!!!

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve exactly for x if possible. If not, find an approximate solution (two decimal places). 3*2^(5x+3)=7*6^x Any help would be greatly appreciated!!!      Log On


   

Question 279680: Solve exactly for x if possible. If not, find an approximate solution (two decimal places).
3*2^(5x+3)=7*6^x
Any help would be greatly appreciated!!!
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
3%2A2%5E%285x%2B3%29=7%2A6%5Ex
With the variables in exponents we usually use logarithms to solve the equation. So we'll start by finding the logarithm of each side. We will get a simpler expression if we choose base 2 or base 6 logarithms. I'll use base 2:
log%282%2C+%283%2A2%5E%285x%2B3%29%29%29+=+log%282%2C+%287%2A6%5Ex%29%29
Next we'll use the properties of logarithms:
  • log%28a%2C+%28p%2Aq%29%29+=+log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29
  • log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29

to split up each logarithm:

Then

This last step, where we moved the exponents of the arguments of the logarithms out in front, is the reason we use logarithms on problems like this. This property of logarithms let's us move the exponent out "where we can get at it".

Now we solve for x. Start by simplifying. First log%282%2C+%282%29%29+=+1 by definition so the second logarithm can be replaced with a 1. And since 2 is a factor of 6, I'm going to use that to my advantage:



Next I'll gather the x terms on one side and the other terms on the other side:

Now I'll factor out the x on the right side:

which simplifies to:
log%282%2C+%283%29%29+%2B+3+-+log%282%2C+%287%29%29+=+x%2A%28log%282%2C+%283%29%29+-+4%29
And last of all, divide both sides by %28log%282%2C+%283%29%29+-+4%29:

which is an exact expression for x.

For a decimal approximation of x, you would have to use the base conversion formula to change the base 2 logarithms into base 10 or base e (ln) logarithms so you could use your calculators.