SOLUTION: I keep messing up these, it's so confusing. Solve a)7^2x+1 = 4^x-2 b) 4^x = 3+18(4^-�) C) 9=logbase5 (x+100) +6

Question 1043601: I keep messing up these, it's so confusing.
Solve
a)7^2x+1 = 4^x-2
b) 4^x = 3+18(4^-�)
C) 9=logbase5 (x+100) +6
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Solve
a)7^2x+1 = 4^x-2
What are the exponents?
----------
b) 4^x = 3+18(4^-�)
Multiply thru by 4^x
4^(2x) = 3*4^x + 18
4^(2x) - 3*4^x - 18 = 0
--
Sub u for 4^x
4u^2 - 3u - 18 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=297 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 2.52921099245176, -1.77921099245176. Here's your graph:

===========
Ignore the negative solution.
4^x =~ 2.52921
x*log(4) = log(2.52921)
x = log(2.52921)/log(4)
x =~ 0.66934
==========================
C) 9=logbase5 (x+100) +6

x+100 = 125
x = 25

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