SOLUTION: Solve for "X" Round to four decimals places. log base 5 (-3x+1)= -4?

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Question 385143: Solve for "X" Round to four decimals places.
log base 5 (-3x+1)= -4?

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%285%2C+%28-3x%2B1%29%29=+-4
With the variable in th e argument of a logarithm, you will often start the solution 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. With this 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 pattern on your equation we get:
-3x%2B1+=+5%5E%28-4%29
TO simplify the right side we need to remember how negative exponents work. 5%5E%28-4%29+=+1%2F5%5E4+=+1%2F625. So our equation is now:
-3x%2B1+=+1%2F625
This is an easy equation to solve. Subtracting 1 from each side we get:
-3x+=+-624%2F625
Divide both sides by -3 we get:
x+=+208%2F625

When solving logarithmic equations like yours, you must check your answers. You must ensure that the arguments (and bases) of any logarithms remain positive. If a solution makes an argument (or base) of any logarithm zero or negative then we must reject that solution. This can happen even if no mistakes were made while solving the equation! This is why we must check.

When checking, always use the original equation:
log%285%2C+%28-3x%2B1%29%29=+-4
Checking x+=+208%2F625:
log%285%2C+%28-3%28208%2F625%29%2B1%29%29=+-4
which simplifies as follows:
log%285%2C+%28-624%2F625%29%2B1%29%29=+-4
log%285%2C+%281%2F625%29%29=+-4
We can see that the argument is positive (just barely) so we have no reason to reject this solution. The rest of the check will tell us if we made a mistake. You are welcome to finish the check.