SOLUTION: logx 1/106=-2

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Question 374413: logx 1/106=-2
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%28x%2C+%281%2F106%29%29=-2
To solve a logarithmic equation like this, where the variable is in the argument (or base) of a logarithm, you often start by 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! So we can proceed from there. With the first form, the next step is to rewrite the equation in exponential form. In general log%28a%2C+%28p%29%29+=+q is equivalent to p+=+a%5Eq. Using this on your equation we get:
1%2F106+=+x%5E%28-2%29
Since x%5E%28-2%29+=+1%2Fx%5E2 this becomes:
1%2F106+=+1%2Fx%5E2
This is a proportion (a fraction = a fraction) so we can use cross-multiplying:
1%2Ax%5E2+=+106%2A1
or
x%5E2+=+106
We can now find the square root of each side:
sqrt%28x%5E2%29+=+sqrt%28106%29
abs%28x%29+=+sqrt%28106%29
x+=+sqrt%28106%29 or x+=+-sqrt%28106%29

And last of all we need to check our answers. This is important with logarithmic equations, not just a good idea. You must make sure that all bases and all arguments of logarithms are positive!

When checking answers you should use the original equation:
log%28x%2C+%281%2F106%29%29=-2
Checking x+=+sqrt%28106%29:
log%28%28sqrt%28106%29%29%2C+%281%2F106%29%29=-2
Both the base and the argument of the logarithm are positive. The rest of the check just determines if our answer works (i.e. whether we made a mistake or not). While I can see that this answer does actually work, maybe you can't. You can finish the check to see.

Checking x+=+-sqrt%28106%29:
log%28%28-sqrt%28106%29%29%2C+%281%2F106%29%29=-2
Already we can see that the base of this logarithm is negative. So we reject this solution because bases (and arguments) of logarithms must always be positive.

So the only solution to
log%28x%2C+%281%2F106%29%29=-2
is
x+=+sqrt%28106%29