Question 413561
{{{2log((7^3))+log((7^(2x)))=log((7^(3x-5)))}}}
Solving equations where the variable is in the argument of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = expression
or
log(expression) = log(expression)<br>
Since your equation is made up of entirely with logarithmic terms, the second, "all-log" form will be easier to achieve. All we have to do is find a way to combine the two logarithms on the left into a single logarithm.<br>
The terms on the left are not like terms so we cannot just add them together. (Like logarithmic terms have the same bases and same arguments.)<br>
Fortunately there is another way to combine logarithmic terms. Two properties of logarithms:<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 the same base and coefficients of 1. Your logarithms have the same base. But the first one does not have a coefficient of 1. Fortunately a third property, {{{q*log(a, (p)) = log(a, (p^q))}}}, allows us to move a coefficient into the argument as its exponent. Using this property on the first logarithmic term we get:
{{{log(((7^3)^2))+log((7^(2x)))=log((7^(3x-5)))}}}
which simplifies to:
{{{log((7^6))+log((7^(2x)))=log((7^(3x-5)))}}}
Now we can use one of the properties to combine the two logarithms. We use the first one because of the "+" between the two logs:
{{{log((7^6*7^(2x)))=log((7^(3x-5)))}}}
which simplifies to:
{{{log((7^(6+2x)))=log((7^(3x-5)))}}}
We finally have the second form. With this form the next step is based on some simple logic. The equation says two base 10 logarithms are equal. The only way that two logarithms of tha same base can be equal is if the arguments are equal. So:
{{{7^(6+2x)=7^(3x-5)}}}
This equation says that two powers of 7 are equal. The only way this can happen is if the exponents are equal, too. So:
6+2x = 3x-5
This is now a very simple equation to solve. Subtracting 2x from each side:
6 = x-5
Adding 5 to each side:
11 = x<br>
Solving logarithmic equations like your requires that answers be checked to make sure that all arguments of logarithms are positive.  Use the original equation to check:
{{{2log((7^3))+log((7^(2x)))=log((7^(3x-5)))}}}
Right away we can tell the no matter what x is, the arguments will be positive. All the arguments in this equation are powers of 7. Powers of 7 can <i>never</i> be zero or negative no matter what the exponent is. So we don't even need to substitute 11 for x to know that the arguments will be positive.