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

[anemone]

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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[anemone]

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:

Spoiler

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 $$a = \frac{\log_2x}{1-\log_23} = \frac1{1-\log_23}.$$