Question

Solve the set of two reals for the following exponential equation:
`\Large\text{${(27^x\:-\:3^x)/(9^x\:-\:3^x) = 82}$}`​

Answer 1

`\sf{(27^x\:-\:3^x)/(9^x\:-\:3^x) = 82}`

`\sf{(9^x(3^x)\:-\:3^x)/(3^x(3^x)\:-\:3^x) = 82}`

`\sf{(3^x(9^x-1))/(3^x(3^x-1)) = 82}`

`\sf{\frac{\cancel{3^x}(9^x-1)}{\cancel{3^x}(3^x-1)} = 82}`

`\sf{((9^x-1))/((3^x-1)) = 82}`

`\sf{((3^x-1)(3^x+1))/((3^x-1)) = 82}`

`\sf{(3^x+1)= 82}`

`\sf{(3^x)= 81}`

`\sf{(3^x)= 3^(4)}`

`\boxed{\bold{x= 4}}`

Answer 2

Answer: The value of the variable x in this exponential equation is 4.

The problem simply asks to solve the following exponential equation on the set of reals:

`\begin{gathered} \bold{(27^x~-~3^x)/(9^x~-~3^x)~=~82}\\\\\\ \bold{(\left(3^3\right )^x~-~3^x)/(\left(3^2\right)^x~-~3^x)~=~82}\\\\\\ \bold{(3^(3x)~-~ 3^x)/(3^(2x)~-~3^x)~=~82}\end{gathered}`

To solve this exponential equation in a much simpler way, what we're going to try to do is factor. Note that if we take
`\bold{ 3^x3}`

x as a common factor what we will get is:

`\bold{(\\ot\!\!3^x\cdot\left(3^(2x)~-~1\right))/(\\ot\!\!3^x\cdot\left( 3^(x)~-~1\right))~=~82}\\\\\\ \bold{ (3^(2x)~-~1)/(3^x~-~1)~=~82}`

We can see that factoring the first part of our exponential equation resulted in a much easier exponential equation to solve. Note that
`\sf 3^(2x)~-~13`

− 1 by the laws of exponents can be written as
`\sf \left(3^x\right)^2~-~1^ 2`

by definition
`\sf x^2~-~y^2~=~\left(x~+~y\right)\cdot\left(x~-~y\right)` so we have:

The exponential equation we now have is not at all difficult to solve. If you repeat the powers of the number 3, you can see that
`\sf 3^4`

Is the same as 81, so we have:

`\begin{gathered} \bold{\\ot\!\!3^x~=~\\ot\!\!3^4} \\\\\\ \boxed{\sf \bold{\therefore~x~=~4}}~~ \bold{\Rightarrow ~Answer}\end{gathered}`