How to use the logarithm method to solve $18^{4x-3}=(54\sqrt{2})^{3x-4}$ for $x$?

Question

What value will satisfy this equation: $$18^{4x-3}=(54\sqrt{2})^{3x-4}$$

Please use the logarithm method. I am having a problem in expressing $54\sqrt{2}$ in the power of $18$. My book simply says that it is $18^{3/2}$. But I can't understand how.

Answer 1

$54\sqrt2=18*3\sqrt2 = 18 *\sqrt18 = 18^\frac{3}{2}$ as the square root is also the positive one half exponent which can added if the terms of multiplied.

Thus, your problem could be rewritten as $4x-3=\frac{3(3x-4)}{2}$ which could be rewritten as $8x-6=9x-12$ which is fairly easy to solve.

Answer 2

1. Hint #1: Take the logarithm of both sides. Think about what base you might want to use.

2. Hint #2: Write $ 54 \sqrt {2} $ in terms of $18$.

3. Hint #3: Let $ 18^x = y $. Rewrite in terms of $y$.

Answer 3

Hint: First, $$ \begin{align} 18^{3/2} &=18\sqrt{18}\\ &=18\cdot3\sqrt2\\ &=54\sqrt2\tag{1} \end{align} $$ Therefore, $\log(54\sqrt2)=\frac32\log(18)$.

Next, take the log of both sides of your equation and use $(1)$.