Find the roots of the given equation : $2^{x+2}.3^{\frac{3x}{x-1}} =9$ - Logarithm problem

Question

Find the roots of the given equation :

$2^{x+2}.3^{\frac{3x}{x-1}} =9$

My working :

Taking log on both sides we get : $$\log (2^{x+2}.3^{\frac{3x}{x-1}}) =\log 3^2 \Rightarrow (x+2)(\log2) + \frac{3x}{x-1}\log 3 = 2\log 3$$

Now how to proceed further in this problem... please suggest thanks....

Answer 1

Think of something to multiply the entire equation by that will remove any $x$'s from the denominator.

Answer 2

As $\log_aa=1$ and $\log_a(a^m)=m\log_aa=m$

taking logarithm wrt $3,$

$$(x+2)\log_32+\frac{3x}{x-1}=2$$

$$\implies(x+2)\log_32=2-\frac{3x}{x-1}=-\frac{x+2}{x-1}$$

$$\implies (x+2)\left(\log_32+\frac1{x+1}\right)=0$$

We know if $a\cdot b\cdot c\cdots=0;$ at least one of $a,b,c,\cdots$ is zero

Answer 3

Hint:

$$2^{x+2}\cdot3^{\frac{3x}{x-1}}=3^2\iff 2^{x+2}=3^{2-\frac{3x}{x-1}}=3^{-\frac{x+2}{x-1}}\iff \begin{cases}x+2=0\\{}\\x-1=-\log_23\end{cases} \;\;(\text{why?)}$$

Answer 4

we could argue that 9 is divisible by 3 and not 2 so wlog

$\large2^{x+2}3^{\frac{3x}{x-1}}=9$

$2^{x+2}$ should be one,so $x=-2$ also

$ \large 3^{\frac{3x}{x-1}}=3^2$

${\frac{3x}{x-1}}=2\implies x=-2$