Question 678478: Find X so that log base 3 of X + log base 3 of (X-3) = log base 3 of 10. Found 2 solutions by nerdybill, jsmallt9:Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! Find X so that log base 3 of X + log base 3 of (X-3) = log base 3 of 10.
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x = {-2, 5}
we can throw out the -2 (extraneous -- won't work) leaving:
x = 5
You can put this solution on YOUR website!
Solving equations where the variable is in the argument of a logarithm often starts by transforming the equation into one of the following general forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)
Since your equation's terms are all logarithms, we will find it easier to transform it into the "all-log" second form.
All we have to do to reach the desired form is to find a way to combine the two logs on the left side. They are not like terms so we cannot just add them together. (Like logarithmic terms have the same bases and the same arguments.) But there is a property of logarithms, , which gives us an alternate way of combining two logs that have a "+" between them. This property requires the same base and coefficients of 1. Our logs meet both requirements so we can use the property:
which simplifies to:
We now have the desired form.
The next step with this form is based on some simple logic. The equation says that two base 3 logs are equal. More precisely, the exponent to put on three to get is the same as the exponent to put on a three to get 10. The only way the exponents can be the same is if the results are the same. So:
Now that the logarithms are gone, we can use "regular" algebra to solve the equation. The equation is quadratic so we want one side to be zero. Subtracting 10 from each side:
Now we factor (or use the Quadratic Formula). This factors fairly easily:
From the Zero Product Property we know that
x - 5 = 0 or x + 2 = 0
Solving these we get:
x = 5 or x = -2
Last, we check. This is not optional! When solving logarithmic equations like this we must check and ensure that all arguments remain positive for each solution. Any "solution" that makes any argument zero or negative must be rejected. Use the original equation to solve:
Checking x = 5:
We can already see that all the arguments will be positive. So this solution checks out.
Checking x = -2:
We can already see that the first two arguments will be negative if x = -2. So this solution fails the check. We must reject it. (Even if only one argument had turned out negative (or zero) then we would still reject the solution.)
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