SOLUTION: Solving exponential equations (2/3)e^4x + (1/3) = 4 I substracted 1/3 from each side but them I'm stuck and don't know what to do (2/3)e^4x = 3(2/3) How do I solve this p

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solving exponential equations (2/3)e^4x + (1/3) = 4 I substracted 1/3 from each side but them I'm stuck and don't know what to do (2/3)e^4x = 3(2/3) How do I solve this p      Log On


   



Question 403941: Solving exponential equations
(2/3)e^4x + (1/3) = 4
I substracted 1/3 from each side but them I'm stuck and don't know what to do
(2/3)e^4x = 3(2/3)
How do I solve this problem?

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%282%2F3%29e%5E%284x%29+%2B+%281%2F3%29+=+4
Isolating the base and its exponent is where you want to start. So subtracting 1/3 is the right first step:
%282%2F3%29e%5E%284x%29++=+11%2F3
To finish isolating the base and its exponent we get rid of the 2/3 by dividing by 2/3 or by multiplying both sides by its reciprocal. Dividing by fractions is a bit of a pain so I'm going to multiply by the reciprocal of 2/3:
%28%282%2F3%29e%5E%284x%29%29%283%2F2%29++=+%2811%2F3%29%283%2F2%29
which simplifies to:
e%5E%284x%29++=+11%2F2
Now that the base and its exponent are isolated, we find the logarithm of each side. The base we use for the logarithm is not really important. But there are two reasons to choose nase e logarithms (better known as ln):
  • e is also the base for the exponent. Using a base e logarithm will result in a simplier expression.
  • Many calculators have a button for ln. This will allow us to find a decimal approximation if needed/desired.

Finding the base e logarithm we get:
ln%28e%5E%284x%29%29++=+ln%2811%2F2%29
Next we use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, to move the exponent of the argument out in front. (It is this very property of logairthms that is the reason we use logarithms on equations like this. Moving the exponent, where the variable is, out in front puts the variable in a place where we can solve for it.)
4x%2Aln%28e%29++=+ln%2811%2F2%29
By definition ln(e) = 1 so this becomes:
4x+=+ln%2811%2F2%29
Last of all we divide by 4 (or multiply by the reciprocal of 4):
%284x%29%281%2F4%29+=+%28ln%2811%2F2%29%29%281%2F4%29
whiich simplifes to:
x+=+%281%2F4%29ln%2811%2F2%29
This is an exact expression for the solution. If you want/need a decimal approximation, get out your calculators. (If your calculator does not have an "ln" button, then you can use the base conversion formula, log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29, to convert to base 10 logs. This would be: x+=+%281%2F4%29%28log%28%2811%2F2%29%29%2Flog%28%28e%29%29%29. And if your calculator has no ln button then its may not have a button for "e", either. In this case use 2.7182818284590451 (or some rounded off version of it) for e: x+=+%281%2F4%29%28log%28%2811%2F2%29%29%2Flog%28%282.7182818284590451%29%29%29