High School What Are the Real Values of $a$ in This Logarithmic Equation?

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The discussion revolves around solving the logarithmic equation $\log_2 (2^{a-1}+3^{a+1})=2a-\log_2(3^a)$. Participants share their methods and solutions, with several members successfully finding the real values of $a$. The solutions involve manipulating the logarithmic expressions and applying properties of logarithms. Key contributors include MarkFL, lfdahl, kaliprasad, and Opalg, who provided correct answers. The thread highlights the collaborative effort in tackling complex mathematical problems.
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Find all real values of $a$ for which $\log_2 (2^{a-1}+3^{a+1})=2a-\log_2(3^a)$.

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Re: Problem of the week #109 -April 28th, 2014

Congratulations to the following members for their correct solutions!:)

1. MarkFL
2. lfdahl
3. kaliprasad
4. Opalg

Solution from Opalg:
Raise 2 to both sides of the equation: $2^{a-1} + 3^{a+1} = \dfrac{2^{2a}}{3^a}$.

Divide both sides of that by $3^a$, getting $\frac12\bigl(\frac23\bigr)^a + 3 = \frac12\bigl(\frac23\bigr)^{2a}$. Now let $x = \bigl(\frac23\bigr)^a$, so that $\frac12x + 3 = x^2.$ That is a quadratic equation with solutions $x = 2,\; -\frac32$. But $x$ cannot be negative, so we must have $x=2$, and $$a = \frac{\log_2x}{1-\log_23} = \frac1{1-\log_23}.$$
 

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