You can put this solution on YOUR website!
Solving equations where the variable is in the argument of a logarithm, like this equation, you often start by transforming the equation into one o fthe following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)
With the "non-log" tgerm of 1 on the right side, the second, all-log form will be more difficult to achieve. So we will aim for the first form. We'll start by gathering the logarithms on one side of the equaiton. Adding to each side we get:
Somehow we need to combine the two logarithms into one. Fortunately there is a property of logarithms, , which will allow us to to combine the logarithms:
We now have the first form.
With the first form, the next step is to rewrite the equaiton in epxonential form. In geneeral is equivalent to . Using this pattern on your equation we get:
We can solvet his equation. First we simplify:
This is a quadratic equation so we want one side to be zero. Subtracting 27 from each side we get:
To simplify the remainder of the solution I will eliminate the decimal by multiplying both sides by 10:
Now we factor (or use the Quadratic Formula). FIrst we factor out the GCF (which is 2):
Next we factor the trinomial:
From the Zero Product Property we know that this (or any) product can be zero only if one (or more) of the factors is zero. So:
2 = 0 or 5x-27 = 0 or x+5 = 0
The first equation has no solution since 2 is never equal to zero. But we can solve the other two equaitons. We should get: or x = -5
When solving logarithmic equaiotns you must check your answers. You must ensure that no base or argument of a logarithm becomes zero or negative. And when checking, always use the original equation:
Checking :
27/5 = 5.4 in decimal form. From this we can see that both arguments will be positive. And the two bases (27) are also positive. The rest of the check will just tell us is we have made an error. (I'll leave that up to you.)
Checking x = -5:
We can already see that both arguments will be negative when x = -5. So we must reject this solution because arguments of logarithms may never be negative (or zero). (If even only one argument turned out negative (or zero) we would still have to reject the solution!)
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