Question

What is the solution to the equation g^(x-2)=27

Answer 1

First problem: Solving for g.

`g=27^{(1)/(x-2)}`

Second problem: Solving for x.

`x=\log_g(27)+2`

Third problem: Assuming g is 9 while solving for x.

`x=3.5`

Explanation:

First problem: Solving for g.

`g^(x-2)=27`

Raise both sides by 1/(x-2).

`(g^(x-2))^{(1)/(x-2)}=27^{(1)/(x-2)}`

`g^(1)=27^{(1)/(x-2)}`

`g=27^{(1)/(x-2)}`

Second problem: Solving for x.

`g^(x-2)=27`

x is in the exponent so we have to convert to logarithm form since we desire to solve for it:

`\log_g(27)=x-2`

Add 2 on both sides:

`\log_g(27)+2=x`

`x=\log_g(27)+2`

Third problem: Assuming g is 9 while solving for x.

`9^(x-2)=27`

I'm going to solve this in a different way than I did above but you could solve it exactly the way I did for x when 9 was g.

I'm going to write both 9 and 27 as 3 to some power.

9=3^2 while 27=3^3.

`(3^2)^(x-2)=3^3`

`3^(2x-4)=3^3`

Since both bases are the same on both sides, we need the exponents to be the same:

`2x-4=3`

Add 4 on both sides:

`2x=7`

Divide both sides by 2:

`x=(7)/(2)`

`x=3.5`

Now earlier for x in terms of g we got:

`x=\log_g(27)+2`

I we input 9 in place of g and put it into our calculator or use some tricks without the calculator to compute we should get 3.5 as the answer like we did above when g was 9.

`x=\log_9(27)+2`

`x=(3)/(2)+2`

`x=1.5+2`

`x=3.5`