SOLUTION: Find the solution of the logarithmic equation: log2 (of base 2)(x^2+ 1x-2)=2

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Question 272460: Find the solution of the logarithmic equation:
log2 (of base 2)(x^2+ 1x-2)=2

Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!

With a variable in the argument of a logarithm, you want the equation in one of the following forms:
log(variable-expression) = other-expression
or
log(variable-expression) = log(other-expression)

Fortunately your equation is already in the first form. To solve an equation in this form we rewrite the equation in exponential form:

or

We now have eliminated the logarithm. And we have a quadratic equation to solve. We want one side to be zero so we'll subtract 4 from each side:

Next we factor the non-zero side (or use the Quadratic Formula). This factors fairly easily:

From the Zero Product Property we know that this product can be zero only if one of the factors is zero. So:
x+3 = 0 or x-2 = 0
Solving these we get:
x = -3 or x = 2

With logarithmic equations it is important, not just a good idea, to check your answers. When checking, one thing to look for are arguments of logarithms that are negative or zero. Arguments of logarithms must never be negative or zero. Any solution that makes an argument negative or zero must be rejected.
Use the original equation when checking.

Checking x = -3:




Rewriting in exponential form:
Check!

Checking x = 2:




Rewriting in exponential form:
Check!

Both solutions check so there are two solutions to your equation: x = -3 or x = 2.

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