Question

Find x in the equation, ${9^{2x}} = {27^{x - 1}}$

Answer 1

Whenever there is equation solving with the variable as an exponent on either of the LHS or the RHS, one must look for equal bases. With equal bases, the equations are simplified better. One more way to solve such an equation is to use logarithmic functions.

Complete step by step solution:
Let us refer,
Left hand side (LHS) =${9^{2x}}$
Right hand side (RHS) = ${27^{x - 1}}$
Upon scrutinizing the LHS and the RHS, we see that the variable “x” is in the exponent parts on both the sides.
In order to equate the sides, we must set the bases of both the LHS and the RHS as equal.
We know that, ${3^2} = 9$ and ${3^3} = 27$.
Putting these values in the LHS and the RHS, we can rewrite the equation as,
${9^{2x}} = {27^{x - 1}} \\\ \Rightarrow {3^{2(}}^{2x)} = {3^{3(}}^{x - 1)} \\\$
Now, we see that both the LHS and the RHS bear the same base, 3.
Since we know that,
${a^b} = {a^{^c}} \\\ \Rightarrow b = c \\\$
Therefore,

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

$\Rightarrow 4x = 3x - 3$
We now have a linear equation to deal with.
Solving this we get,
$\Rightarrow x = - 3$

Note: An exponential equation is one with a term that has an exponent greater than one. For example, ${x^{3/2{\text{ }}}} + {\text{ }}2x{\text{ }} + {\text{ }}1$ is an expression while $2x{\text{ }} + {\text{ }}3\;$isn't an expression. Similarly, ${x^3} = 27$ is an equation while $x{\text{ }} + {\text{ }}2{\text{ }} = {\text{ }}29\;$ isn't an equation. Base is the number that is multiplied by itself a certain quantity of times. Exponent is the number of times a quantity is multiplied by itself.