Solving logarithm equation help?

[liz777]

Homework Statement

log₂(x+2)=log₂x²

-log base 2(X+2)= log base 2 (x²)

The Attempt at a Solution

I know the answer is supposed to be -1 and 2, but I get the wrong answer every time I try.
I tried bringing log₂x² over to the other side and then got log₂(2/x) which got me nowhere!

What am I supposed to do to solve this? Any tips would be greatly appreciated :)

 

[J Hill]

There is a technique called exponentiation (I think) used to solve these types of equations. Just put both sides of the equation in an exponent that is the same as the logarithm base, like this:
$$2^{log_{2} (x+2)} =2^{log_{2} (x^2}}$$
Then you can use the rules for logs an exponents to get to
x+2 = x^2

 

[Chaos2009]

Right, it is called exponentiation. You can also see that both sides are ultimately a value within the log function. If the log base 2 of both sides is the same, then both sides must be the same. There is another step though that many people forget. When you exponentiate, you may get answers which do not work because on of the log functions is undefined for that value of x. You have to check each value you find for x to make sure that both log functions are defined for that value of x. If you do this particular problem right, both solutions work.

 

[Wretchosoft]

Exponentiation isn't even necessary in this case. log is 1-1, so equality can only hold if the arguments are equal.

 

[Chaos2009]

Well, yes, it's still exponentiating, you're just skipping the steps in between.

 

[th3chemist]

log2(x+2)=log2x2

It's quite easy. The logarithms have the same base. Thus you can just work with it.

(x+2) = x²

Now, set the variable to 0 and factor
x²-x-2 = 0
(x-1)(x+2) = 0

two answers are: x -1 = 0; x=1 or x+2=0; x=-2

:)

 

[liz777]

thanks everyone! this really helped :) Turned out I completely forgot about factoring...