Question 238104
{{{ln(x) + 2*ln(4) = ln(7) - ln(x)}}}
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)<br>
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.<br>
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:<ul><li>{{{log(a, (p)) + log(a, (q)) = log(a, (p*q))}}}</li><li>{{{log(a, (p)) - log(a, (q)) = log(a, (p/q))}}}</li></ul>
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: {{{p*log(a, (q)) = log(a, (p^q))}}}.
So next we will deal with the coefficient on the second term on the left side:
{{{ln(x) + ln(4^2) = ln(7) - ln(x)}}}
which simplifies to:
{{{ln(x) + ln(16) = ln(7) - ln(x)}}}
Now we can use the first two properties to combine the logs on each side:
{{{ln(x*16) = ln(7/x)}}}
{{{ln(16x) = ln(7/x)}}}
And we have achieved the desired form (#2). The next step uses some basic logic:<ul><li>16x is some number and 7/x is some number.</li><li>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.</li><li>Since their logarithms are equal, 16x and 7/x are equal.</li></ul>
So
{{{16x = 7/x}}}
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:
{{{16x^2 = 7}}}
Divide by 16:
{{{x^2 = 7/16}}}
Square root of each side:
{{{sqrt(x^2) = sqrt(7/16)}}}
{{{abs(x) = sqrt(7)/sqrt(16)}}}
{{{abs(x) = sqrt(7)/4}}}
{{{x = sqrt(7)/4}}} or {{{x = -sqrt(7)/4}}}<br>
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 {{{x = sqrt(7)/4}}}
{{{ln(sqrt(7)/4) + 2*ln(4) = ln(7) - ln(sqrt(7)/4)}}}
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 {{{x = -sqrt(7)/4}}}
{{{ln(-sqrt(7)/4) + 2*ln(4) = ln(7) - ln(-sqrt(7)/4)}}}
We can see that the arguments where we substituted for x are negative. So we must reject this solution.<br>
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.