MATH SOLVE

4 months ago

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

Accepted Solution

A:

Answer:First problem: Solving for g.[tex]g=27^{\frac{1}{x-2}}[/tex]Second problem: Solving for x.[tex]x=\log_g(27)+2[/tex]Third problem: Assuming g is 9 while solving for x.[tex]x=3.5[/tex]Step-by-step explanation:First problem: Solving for g.[tex]g^{x-2}=27[/tex]Raise both sides by 1/(x-2).[tex](g^{x-2})^{\frac{1}{x-2}}=27^{\frac{1}{x-2}}[/tex][tex]g^{1}=27^{\frac{1}{x-2}}[/tex][tex]g=27^{\frac{1}{x-2}}[/tex]Second problem: Solving for x.[tex]g^{x-2}=27[/tex]x is in the exponent so we have to convert to logarithm form since we desire to solve for it:[tex]\log_g(27)=x-2[/tex]Add 2 on both sides:[tex]\log_g(27)+2=x[/tex][tex]x=\log_g(27)+2[/tex]Third problem: Assuming g is 9 while solving for x.[tex]9^{x-2}=27[/tex]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.[tex](3^2)^{x-2}=3^3[/tex][tex]3^{2x-4}=3^3[/tex]Since both bases are the same on both sides, we need the exponents to be the same:[tex]2x-4=3[/tex]Add 4 on both sides:[tex]2x=7[/tex]Divide both sides by 2:[tex]x=\frac{7}{2}[/tex][tex]x=3.5[/tex]Now earlier for x in terms of g we got:[tex]x=\log_g(27)+2[/tex]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.[tex]x=\log_9(27)+2[/tex][tex]x=\frac{3}{2}+2[/tex][tex]x=1.5+2[/tex][tex]x=3.5[/tex]