SOLUTION: I have been told I need to turn my logs into exponents to solve this equation but I am not sure how it works. I will do it as far as I know then please help me. I also need to know
Question 847130: I have been told I need to turn my logs into exponents to solve this equation but I am not sure how it works. I will do it as far as I know then please help me. I also need to know how when and why logs are used as exponents so I can solve future equations on my own. Here it is;
Base 3 (I don't have subscript), log3 (x+1) - 2 = log3 (x).
Here is what I have
Move the 2 over to the right side
Log3 (x+3) = log3 (x) + 2
Divide both sides by "x"
Log3 (x+1)/x = log3 (x)/x+ 2
Turn the logs into exponents of the base
3^log3 (x+1) = 3^log3 (x)/x +2 (the 2 is not an exponent)
The base and log bases cancel out and you have
Log3 (x+1)/x = 3^2
Log3 (x+1)/x = 9
Multiply both sides by x/x to get rid of the denominator and that leaves
X+1 = 9x
X + 1 - x = 9x - x
1 = 8x
Divide by 8
1/8 = x
If this is correct, why did the 2 become an exponent?
applied 2nd
Applied 1st