.
The original equation is
= .
Note that the domain, where both sides are defined, is the set of positive real x: { x | x > 0 }.
Due to properties of logarithms, it is the same as
= , or, equivalently,
= , which, in turn, is equivalent to
= .
It implies
x^2 = x + 12
x^2 - x - 12 = 0,
(x-4)*(x+3) = 0.
Of the two roots, x= 4 and x= -3, only positive x= 4 is in the domain and is, therefore, the solution.
ANSWER. x= 4.
Solved. // I mean, solved in a way as it should be done.
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On the way, I used standard and elementary properties of logarithms that every student must learn and must know
before starting solving such problems.
On logarithms and their properties, see the lessons
- WHAT IS the logarithm
- Properties of the logarithm
- Change of Base Formula for logarithms
- Solving logarithmic equations
in this site.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Logarithms".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.