Question

How do you solve \[{9^{2x - 1}} = {27^x}\] ?

Answer 1

Hint: We have a simple algebraic equation. Here we need to solve for ‘x’. We use the concept that when the bases are the same in the equation we can substitute the exponents, that is \[{x^a} = {x^b}\] then we can write \[a = b\] . Here we can express 27 and 9 in terms of 3. Then we apply the exponent law (laws of indices).

Complete step-by-step answer:
Here we have,
 \[{9^{2x - 1}} = {27^x}\]
We can write 9 as \[9 \Rightarrow 3 \times 3 = {3^2}\]
We can write 27 as \[27 \Rightarrow 3 \times 3 \times 3 = {3^3}\]
 \[{\left( {{3^2}} \right)^{2x - 1}} = {\left( {{3^3}} \right)^x}\]
Now applying bracket law that is \[ \Rightarrow {\left( {{x^m}} \right)^n} = {x^{m \times n}}\] ,
 \[{\left( 3 \right)^{2\left( {2x - 1} \right)}} = {\left( 3 \right)^{3x}}\]
 \[{\left( 3 \right)^{4x - 2}} = {\left( 3 \right)^{3x}}\]
We can see that the bases are same then we can substitute the exponents or power, that is
 \[4x - 2 = 3x\]
 \[4x - 3x = 2\]
 \[ \Rightarrow x = 2\] . This is the required answer.
So, the correct answer is “x = 2”.

Note: We can check whether the obtained answer is correct or not. Substitute the obtained answer in the given problem,
 \[{9^{2(2) - 1}} = {27^{(2)}}\]
 \[{9^{4 - 1}} = {27^2}\]
 \[{9^3} = {27^2}\]
 \[ \Rightarrow 729 = 729\] . Hence the obtained answer is correct.