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I am struggling to solve this question, Suppose $$2^x = 6$$, what is the value of $$2^{3x-1}$$ ?

I simply do not know where to start. Please note that I am expected to do this with knowledge up to Algebra 1 for this problem was found in an AMC 8 / Mathcounts Prep book.
NoChance
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Jamlandia
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2 Answers2

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We're looking for $2^{(3x-1)}$ in terms of $2^x$. Calculators are not allowed on the AMC 8 and finding the exact value of the logarithm would be hard to do with algebra 1 knowledge.

Recall that $2^{(3x-1)}$ is the same as $2^{(3x)}*2^{(-1)}$ since multiplying numbers of a common base yields that common base raised to the sum of the exponents, which in this case is $3x-1$. Notice again, that we can rewrite $2^{(3x)}$ as ${2^x}^{(3)}$. Now, we have in all, $${2^x}^{(3)}*2^{(-1)}$$

Just apply the fact that $2^{x}=6$ here and substitute. We are left with $6^3*2^{-1}$. This is equal to $216*.5$ which is $108$.

Gnumbertester
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You have $$ 2^{3x-1}=\frac{(2^x)^3}{2}=\frac{6^3}{2}=108 $$

Tuvasbien
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