SOLUTION: What am I doing wrong here or am I on the right track? log5x - log5(x - 2) = log5 4 log5x - log5 (x - 2) = log5/log4 log5 x-(x-2) = 1.161 -x^2 + 2x - 1.161 = 0 -b +-

Algebra.Com
Question 691014: What am I doing wrong here or am I on the right track?
log5x - log5(x - 2) = log5 4
log5x - log5 (x - 2) = log5/log4
log5 x-(x-2) = 1.161
-x^2 + 2x - 1.161 = 0
-b +- sqrt b^2 - 4(a)(c) / 2(a)
-2 +- sqrt 2^2 - 4(-1)(-1.161) / 2(-1)

Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
Unfortunately, you are far from the right track, and trouble rendering the notation does not help.
I believe the problem is
For such a problem, you should know and use the definition and properties of logarithms (see below).

THE SOLUTION:
Using knowledge about logarithms:
--> -->
Then,
--> --> --> -->
--> --> --> -->

DEFINITION:
Logarithm of a number is defined (in down to earth terms) as the exponent you need to put on the base to get that number.
and
because and
is not such a simple number.
You cannot write its exact value as a fraction or a decimal, but it's approximately 0.8614.

PROPERTTIES OF LOGARITHMS:
Some teachers may give you formulas for the properties of logarithms and expect you to remember them.
However, it is easy to remember (or rediscover) most of those properties by thinking from the definition of logarithm.
SUM OF LOGARITHMS:
When using the same base, multiplying powers means adding the exponents
(example ).
So, when logarithms of the same base are added, you get the logarithm of the product.

DIFFERENCE OF LOGARITHMS:
Similarly, subtracting logarithms of the same base gives you the logarithm of the quotient.

CHANGE OF BASE:
That is not quite so obvious.
To calculate the approximate value of the logarithm of a number in a strange base (like 5), you divide the base 10 logarithms of the number by the base 10 logarithms of the strange base.

So
and since the base can be omitted is it is , you can write it as

RELATED QUESTIONS

Solve:... (answered by lwsshak3)
log5x = log5 (x+4) -2 (answered by stanbon)
log5 x- log5... (answered by user_dude2008)
Solve: log5x-log5(x-2)=log5(4). all the 5s are small. (answered by ankor@dixie-net.com)
log5(x-1)=4+log5(x-2) (answered by nerdybill)
log5 (x+2)-log5 10=log5... (answered by nerdybill)
what is the solution set for... (answered by stanbon)
I am trying to figure out how to solve log5(x+1)-log5x=2 can you... (answered by ewatrrr)
log5(x-2)= 2 + log5(x-4) (answered by lwsshak3)