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Find the value of $x$ satisfying $18^{4x-3}=(54\sqrt{2})^{3x-4}$. The given options are $2,6,3,4$. I don't know how to do this.

Math Lover
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Yash Pranesh
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  • In addition to the other answers, it is reasonable to just plug in the four options and check if they give a true equality if not. – Zach Boyd Sep 09 '17 at 18:35

2 Answers2

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Hint:

$$18 = 2\cdot3^2, \text { and } 54\sqrt{2}=2^{3/2}\cdot3^3.$$

Math Lover
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Notice that, following Math Lover's hint, $\displaystyle 54\sqrt{2}=18^\frac{3}{2}$, because $54=(3\sqrt{2})^3$ and $18=(3\sqrt{2})^2$.

Therefore, $18^{4x-3}=(18^\frac{3}{2})^{3x-4}$, so $18^{4x-3}=18^{\frac{3}{2}(3x-4)}$.

Taking the base $18$ logarithm of each side gives

$ 4x-3=\frac{3}{2}(3x-4)$

$8x-6=9x-12$

$\boxed{x=6}$.

Saketh Malyala
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  • When you see problems like this, first try to establish a connection between the bases, then you can turn the problem into something much easier! – Saketh Malyala Sep 09 '17 at 18:33