using logarithms to solve the following equation to find x

Question

$9^{2x} = 27^{1-x}$ ?? I'm really struggling with this questions. I appreciate your help and if you can please show me your working out so I can understand it too,

Answer 1

Assuming it is this $$9^{2x} = 27^{1-x}$$ then you can use $$ 9 = 3^2\,\,\text{and}\,\,27=3^3 $$ to get $$ \left(3^2\right)^{2x} = \left(3^3\right)^{1-x}\\ 3^{4x}=3^{3(1-x)} $$ you can solve without logs. i.e. $$ 3^a = 3^b\,\,\text{iff}\,\,a=b $$

Answer 2

As $m\log(a)=\log(a^m),$ where both logarithm remain defined.

Taking logarithm in both sides, $$2x\cdot\log(9)=(1-x)\log(27)$$

Now $\log(27)=\log(3^3)=3\log3$ and $\log(9)=\log(3^2)=2\log3$

and $\log3\ne0$ can be cancelled safely from both sides

Answer 3

Noting that $9^{2x}=(3^2)^{2x}=3^{4x}$ and $27^{1-x}=(3^3)^{1-x}=3^{3(1-x)}$ we get $$3^{4x}=3^{3(1-x)}$$ and by taking the logarithm in base $3$ we have $4x=3(1-x)$ whose unique solution is $x=\frac{3}{7}$

Answer 4

$log_3\left( 9^{2x} \right) = log_3\left( 27^{1-x} \right)$

${2x} \cdot log_3\left( 9 \right) = ({1-x}) \cdot log_3\left( 27 \right)$

${2x} \cdot log_3\left( 3^2 \right) = ({1-x}) \cdot log_3\left( 3^3 \right)$

${2x} \cdot 2 = ({1-x}) \cdot 3$

${4x} = 3 - {3x}$

${7x} = 3$

$x = \frac{3}{7}$