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Solving equations where the variable is in the argument (or base) of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression
Your equation is already in the first form! With the first form the next step is to rewrite the equation in exponential form. In general is equivalent to . Using this pattern on your equation we get:
which simplifies to:
Now we can solve for x. First let's get rid of the square root by squaring both sides:
Since there is no "x" term we can just add 1:
and find the square roots:
(NOTE: Algebra.com's software will not let me use a "plus or minus" symbol wiout something in front of it. This is why I put the zero in front. Mathematically the zero is unnecessary.)
With this problem we have two reasons a check is required, not optional:
Any equation where the variable is in the argument of a log must be checked to make sure that the arguments are all positive.
Any time both sides of an equation are squared you must check for extraneous solutions (solution that work in the squared equation but not in the original equation.
Use the original equation to check:
Checking
Simplifying...
Since square roots are positive, we can already see that this solution passes the first check. But we should continue to make sure this solution is not extraneous.
2 = 2 Check!