Question 604207
{{{log((sqrt(x^2-1)))=2}}}
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<br>
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 {{{log(a, (p)) = q}}} is equivalent to {{{a^q = p}}}. Using this pattern on your equation we get:
{{{10^2 = sqrt(x^2-1)}}}
which simplifies to:
{{{100 = sqrt(x^2-1)}}}
Now we can solve for x. First let's get rid of the square root by squaring both sides:
{{{10000 = x^2-1}}}
Since there is no "x" term we can just add 1:
{{{10001 = x^2}}}
and find the square roots:
{{{0 +- sqrt(10001) = x}}}
(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.)<br>
With this problem we have two reasons a check is required, not optional:<ul><li>Any equation where the variable is in the argument of a log must be checked to make sure that the arguments are all positive.</li><li>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.</li></ul>
Use the original equation to check:
{{{log((sqrt(x^2-1)))=2}}}
Checking {{{x = sqrt(10001)}}}
{{{log((sqrt((sqrt(10001))^2-1)))=2}}}
Simplifying...
{{{log((sqrt(10001-1)))=2}}}
{{{log((sqrt(10000)))=2}}}
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.
{{{log(100)))=2}}}
2 = 2 Check!<br>
Checking {{{x = -sqrt(10001)}}}
{{{log((sqrt((-sqrt(10001))^2-1)))=2}}}
Simplifying...
{{{log((sqrt(10001-1)))=2}}}
{{{log((sqrt(10000)))=2}}}
{{{log(100)))=2}}}
2 = 2 Check!<br>
So both {{{sqrt(10001)}}} and {{{-sqrt(10001)}}} are solutions to your equation.