Question

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

Answer 1

Answer:

First problem: Solving for g.

$g=27^{\frac{1}{x-2}}$

Second problem: Solving for x.

$x=\log_g(27)+2$

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

$x=3.5$

Step-by-step explanation:

First problem: Solving for g.

$g^{x-2}=27$

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

$(g^{x-2})^{\frac{1}{x-2}}=27^{\frac{1}{x-2}}$

$g^{1}=27^{\frac{1}{x-2}}$

$g=27^{\frac{1}{x-2}}$

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:

$x=\frac{7}{2}$

$x=3.5$

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.

$x=\log_9(27)+2$

$x=\frac{3}{2}+2$

$x=1.5+2$

$x=3.5$