SOLUTION: Here is the problem I am having trouble with: -Find the solution of the exponential equation, correct to four decimal places. 5^x = 4^x+1 so far I got: ln5^x = ln4^x

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Here is the problem I am having trouble with: -Find the solution of the exponential equation, correct to four decimal places. 5^x = 4^x+1 so far I got: ln5^x = ln4^x      Log On


   

Question 81530This question is from textbook College Algebra
: Here is the problem I am having trouble with:
-Find the solution of the exponential equation, correct to four decimal places.
5^x = 4^x+1

so far I got:
ln5^x = ln4^x+1
(x)ln5 = (x+1)ln4
(x)ln5 = xln4 + ln4
If this is right, I'm not sure how to isolate x at this point.


This question is from textbook College Algebra

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
It looks right so far!!

5^x = 4^x+1
so far I got:
ln5^x = ln4^x+1
(x)ln5 = (x+1)ln4
(x)ln5 = (x)ln4 + ln4

Now, you have to get all the x terms on one side, by subtracting (x) ln4 from each side:
(x) ln5 - (x) ln4= ln4

Factor out the x:
x(ln5-ln4) = ln 4

Divide both sides by (ln5-ln4):
%28x%28ln5-ln4%29%29%2F%28%28ln5-ln4%29%29=+%28ln4%29%2F%28%28ln5-ln4%29%29+

This calculates to : x = 6.2126

If you are using a graphing calculator to calculate this, be sure you know where to put parentheses in the calculator to make it do what it needs to do.

R^2 at SCC