SOLUTION: solve the equation log8(3x+7)=log8(7x-9) log4(x)=3/2 loga(1/8)=-3 log8(x+9)=log8(3x-14) log4(x^2+6)=log4(5x) logy16=4 log7n=-1/2 log7(8x+20)log7(x+6) log5(x^2-30)=log5(6)

Question 594545: solve the equation log8(3x+7)=log8(7x-9)
log4(x)=3/2
loga(1/8)=-3
log8(x+9)=log8(3x-14)
log4(x^2+6)=log4(5x)
logy16=4
log7n=-1/2
log7(8x+20)log7(x+6)
log5(x^2-30)=log5(6)
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
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.

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)

To solve the equations of the form
log(expression) = other-expression
you start by rewriting the equation in exponential form. In general

is equivalent to

For example, your equation

should be rewritten in exponential form:

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:

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):

and of course so we have:

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:

Note how I've removed the "-" since the reciprocal has been done. And 8/1 is just an 8 so now we have:

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.

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:

We just write:

The logic behind this is: The only way the base 4 log of can be equal to the base 4 log of is if the two arguments are equal.

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:

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

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".

Checking the solution a = 2 for the equation :

The base is 2 (positive) and the argument is 1/8 (also positive) so this solution checks out.

Checking the solution x = 2 for the equation :

which simplifies to:

The bases are 4's and the argument are 10's, all positive, so this solution checks out.

Checking the solution x = 3 for the equation :

which simplifies to:

The bases are 4's and the argument are 15's, all positive, so this solution checks out, too

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.
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