How do you solve #log_12 (p^2-5p)=log_12 (8+2p)#?

2 Answers
Larry Leitch-Casey
Apr 16, 2016

Hi there! To solve this, you must recognize that if you have logs of the same base, you can drop them, leaving you with the functions inside.

Explanation:

When you have:

#log_b(f(x)) = log_b(g(x))# this is equivalent to:

# f(x) = g(x) #

Since both logs in your question have a base of 12, you can drop them, leaving you with:

# p^2 -5p = 8 + 2p #

Rearranging we get:

#p^2 - 5p -2p - 8 = 0 #

Collecting like terms:

#p^2 - 7p - 8 = 0#

Factoring this simple trinomial (finding numbers that multiply to -8 and add to -7):

# (p-8)(p+1)=0 #

Solving each piece we get:

# p=8,-1 #

And that's it, those are the p values that would make those expressions equal. Hopefully everything was clear and helpful! If you have any questions, please feel free to ask! :)

A. S. Adikesavan
Apr 16, 2016

#p=-1 and p = 8#.

Explanation:

Both the logarithms have the same base 12.
So, #p^2-5p=8+2p#.
#p^2-7p-8=0#
The roots of this quadratic equation are# -1 and 8.#.
Both the roots are admissible for both the logarithms..

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