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When you are solving for variables that are in the arguments of logarithms you want to use Algebra and/or properties of logarithms to transform the equation into one of the following forms:
log(expression-with-variables) = some-other-expression
or
log(expression-with-variables) = log(some-other-expression)
Since your equation is made up entirely of logarithmic terms, we will work towards the second form. It is a little easier and faster than the first form.
So on each side of the equation we want to "combine" the two logarithms into one. The properties of logarithms that we can use for this are:
These properties require that the coefficients of the logs be 1's (i.e. invisible). And we have a property for dealing with coefficients which are not 1'a: .
So next we will deal with the coefficient on the second term on the left side:
which simplifies to:
Now we can use the first two properties to combine the logs on each side:
And we have achieved the desired form (#2). The next step uses some basic logic:
16x is some number and 7/x is some number.
This equation says that the natural logs of these two numbers are equal. In other words, the exponent for e that results in 16x is the same exponent for e that result in 7/x.
Since their logarithms are equal, 16x and 7/x are equal.
So
Now we have an equation where the variable is no longer in the argument of the logarithm. We can solve this. Start by multiplying by x to eliminate the fraction:
Divide by 16:
Square root of each side: or
With logarithmic equations it is important (not just a good idea) to check your answers. We need to ensure that the possible solutions make the arguments to the logarithms are positive. (Logarithms cannot have zero or negative arguments!)
Checking
We can see that all the arguments are positive. So we do not need to reject this solution. You can complete the check with your calculator.
Checking
We can see that the arguments where we substituted for x are negative. So we must reject this solution.
Note: Do not just reject negative values for x without checking. What makes a solution invalid is not that it is negative. It is the fact that a value makes a logarithmic argument zero or negative that is the problem. Sometimes negative values for x work just fine. Sometimes positive values for x do not work. You
just have to check to see what works and what doesn't.
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