Question 594545
First of all, this is far too many problems for a single post. So I am going to provide mostly general help on these problems.<br>
These equations can be separated into two groups. Those in the form: log(exprssion) = other-expression
and those in the form
log(expression) = log(other-expression)<br>
To solve the equations of the form
log(expression) = other-expression
you start by rewriting the equation in exponential form. In general
{{{log(a, (p)) = q}}}
is equivalent to
{{{a^q = p}}}<br>
For example, your equation
{{{log(a, (1/8)) = -3}}}
should be rewritten in exponential form:
{{{a^(-3) = 1/8}}}
Now you solve the new equation. You are looking for the number which is equal to 1/8 when it is raised to the -3 power. If the answer is not clear to you, then think of the exponent this way: The "-" in an exponent means reciprocal (or upside down) and the 3 means cubed. So what number, upside down and cubed is equal to 1/8? If you still cannot figure out the answer, then raise each side of the equation to whatever power will make the exponent on "a" turn into a 1. In this case it would be to the -1/3 power:
{{{a^(-3)^(-1/3) = (1/8)^(-1/3)}}}
On the left side the rules for exponents tell us to multiply the exponents. And when you multiply -3 and -1/3 you get 1 (which was the whole point of doing this):
{{{a^1 = (1/8)^(-1/3)}}}
and of course {{{a^1 = a}}} so we have:
{{{a = (1/8)^(-1/3)}}}
Now we just have to figure out what the right side works out to be. Again it helps to look at the exponent in parts. The "-" still means reciprocal (or upside down) and the 1/3 means cube root. Note: you can do these in any order and it won't matter. So pick the order that makes it easiest for you. To me flipping 1/8 upside down looks easier than the cube root of 1/8 so I will start with that:
{{{a = (8/1)^(1/3)}}}
Note how I've removed the "-" since the reciprocal has been done. And 8/1 is just an 8 so now we have:
{{{a = 8^(1/3)}}}
We have just the cube root left. So the cube root of 8 is the answer to this problem. What number cubed is equal to 8? (Not: "What is 8 cubed?). If you still don't know the answer, enter 8^(1/3) into your calculator. Once you have the answer, look back and see if you can understand how you might have figured this out earlier.<br>
The equations of the form
log(expression) = log(other-expression)
are easier. (NOTE: The two logs MUST have the same base for this to work!) To solve just set the two arguments equal and solve. For example:
{{{log(4, (x^2+6))=log(4, (5x))}}}
We just write:
{{{x^2+6 = 5x}}}
The logic behind this is: The only way the base 4 log of {{{x^2+6}}} can be equal to the base 4 log of {{{5x}}} is if the two arguments are equal.<br>
Now we just solve the new equation. Since this is a quadratic equation we want one side to be zero. Subtracting 5x from each side we get:
{{{x^2 - 5x + 6 = 0}}}
Now we factor (or use the Quadratic Formula). This factors pretty easily:
(x -2)(x-3) = 0
The Zero Product Property tells us that the only way this (or any) product to be zero, one of the factors must be zero. So
x - 2 = 0 or x - 3 = 0
Solving these we get:
x = 2 or x = 3<br>
IMPORTANT: When solving logarithmic equations (either form) you must check your answer(s)! You must ensure that all bases and arguments of the logs are positive. If a "solution" makes any base and/or any argument of a log zero or negative you must reject that "solution".<br>
Checking the solution a = 2 for the equation {{{log(a, (1/8)) = -3}}}:
{{{log(2, (1/8)) = -3}}}
The base is 2 (positive) and the argument is 1/8 (also positive) so this solution checks out.<br>
Checking the solution x = 2 for the equation {{{log(4, (x^2+6))=log(4, (5x))}}}:
{{{log(4, ((2)^2+6))=log(4, (5(2)))}}}
which simplifies to:
{{{log(4, (10))=log(4, (10))}}}
The bases are 4's and the argument are 10's, all positive, so this solution checks out.<br>
Checking the solution x = 3 for the equation {{{log(4, (x^2+6))=log(4, (5x))}}}:
{{{log(4, ((3)^2+6))=log(4, (5(3)))}}}
which simplifies to:
{{{log(4, (15))=log(4, (15))}}}
The bases are 4's and the argument are 15's, all positive, so this solution checks out, too<br>
One last thing. If a logarithmic equation is not in one of these two forms, solving will usually start with transforming it into one of these forms.