Solving Logarithmic Equations: Step-by-Step Guide

[majormuss]

Homework Statement

My book has these steps for evaluating a logarithmic equation and I don't quite understand what is going on. It says...
log (2) x(x-4)=5
x(x-4)=2^5 <---------> this is the level I am getting problems... how does log(2) divide the 5 on the other side and end up getting 2^5 ??
By the way this is how the equation continues( which I understand)
X^2-4x -32
(x-8)(x+4)= 0
x=8 x=-4

Homework Equations

The Attempt at a Solution

 

[Dick]

If log_2(a)=b then a=2^b. It's sort of the definition of log.

 

[HallsofIvy]

Homework Statement

My book has these steps for evaluating a logarithmic equation and I don't quite understand what is going on. It says...
log (2) x(x-4)=5
x(x-4)=2^5 <---------> this is the level I am getting problems... how does log(2) divide the 5 on the other side and end up getting 2^5 ?

It doesn't! "[itex]log_2 x[/itex] does NOT mean "[itex]log_2[/itex] multiplied by x" and so you are not "dividing by [itex]log_2[/itex]"

[itex]log_2 x [/itex] means "apply the function "logarithm base 2" to x. You remove that function by applying the inverse function to both sides. And the inverse function to [itex]f(x)= log_2 x[/itex] if [itex]f^{-1}(x)= 2^x[/itex]. Specifically, [itex]f(f^{-1}(x))= x[/itex] and [itex]f^{-1}(f(x))= 2^{log_2 x}= x[/itex].

Applying the inverse function of [itex]log_2(x)[/itex], [itex]2^x[/itex], to both sides:
[tex]2^{log_2(x(x-4))}= 2^5[/tex]
[tex]x(x-4)= 2^5[/tex].

What did you learn as the definition of "[itex]log_2(x)[/itex]"?

By the way this is how the equation continues( which I understand)
X^2-4x -32
(x-8)(x+4)= 0
x=8 x=-4

Homework Equations

The Attempt at a Solution