SOLUTION: How do you algebraically find the solution to the following problem? log(79-3x) + log(x) = 2

Question 195310: How do you algebraically find the solution to the following problem?
log(79-3x) + log(x) = 2
Found 2 solutions by ankor@dixie-net.com, solver91311:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
How do you algebraically find the solution to the following problem?
log(79-3x) + log(x) = 2
:
Terms are multiplied when logs are added, we can write it:
log(x(79-3x)) = 2
:
log(79x - 3x^2) = 2
:
Find the 10^x of both sides:
79x - 3x^2 = 100:
:P
Arrange as a quadratic eq on the right:
0 = 3x^2 - 79x + 100
:
This factors to:
(3x - 4)(x - 25) = 0
:
Solutions:
x = 25
and
x =
;
;
Check solutions in original problems
x=25
log(79-3(25)) + log(25) = 2
log(79-75) + log(25) = 2
log(4) + log(25) = 2
.602 + 1.398 = 2
and
x=1.33
log(79-3(1.33)) + log(1.33) = 2
log(79-4) + log(1.33) = 2
log(75) + log(1.33) = 2
1.875 + .124 ~ 1.99 ~ 2

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

The sum of the logs is the log of the product, i.e.

, so:

, so:

But:

, so:

Just solve the quadratic. Hint: It factors.

John

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