SOLUTION: how do i solve (log base 4 (x+3))+(log base 4 (x-3))=2?

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Question 390652: how do i solve (log base 4 (x+3))+(log base 4 (x-3))=2?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log base 4 (x+3))+(log base 4 (x-3))=2
----
log4(x+3) + log4(x-3) = 2
-------
log4[x^2-9] = 2
---
base = 4
exponent = 2
result = x^2-9
---
Write in exponential form:
x^2-9 = 4^2
---
x^2 = 16+9
---
x^2 = 25
---
Positive solution:
x = 5
=======================
Cheers,
Stan H.

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


I have no idea how you solve



I have no idea how you do anything that you do.

However, on the outside chance that the way I solve it is of value to you, this is the way I would proceed:

The sum of the logs is the log of the product, so:





The definition of the logarithm function is:



which we can use to write:



which becomes:



Solve the quadratic. Don't forget to specify both roots.

By the way, just out of idle curiousity, is the personal pronoun, I, rendered in your post in lower case because you are lazy or because you are submissive?

John

My calculator said it, I believe it, that settles it
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