Question 1105066
the log rules that apply are:


log(x) + log(y) = log(x * y)


log(x) - log(y) = log(x / y)


if log(x) = log(y), then x = y


when you solve this problem, you use these properties of logs.


start with log2(5)+log2(x+2)-log2(x+1)=log2(7)


since log2(5)+log2(x+2) = log2(5*(x+2)), then your equation becomes:


log2(5*(x+2)) - log2(x+1) = log2(7)


since log2(5*(x+2)) - log2(x+1) = log2(5*(x+2)/(x+1)), then your equation becomes:


log2(5*(x+2)/(x+1)) = log2(7)


this is true if and only if 5*(x+2)/(x+1) = 7


multiply both sides of this equation by (x+1) to get:


5*(x+2) = 7*(x+1)


simplify by distributing the multiplication to get:


5x+10 = 7x+7


subtract 5x from both sides of the equation and subtract 7 from both sides of the equation to get:


10-7 = 7x-5x


combine like terms to get:


3 = 2x


solve for x to get x = 2/3.


that's your solution.


confirm by replacing x in the original equation with 2/3 to confirm that the original equation is true with that value of x.


the following tutorials contain good info on logs.


read the tutorials and do the exercises.
that will help.


<a href= "http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm" target = "_blank">http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm</a>


here's more from a different source.


<a href = "http://www.themathpage.com/aprecalc/logarithms.htm" target = "_blank">http://www.themathpage.com/aprecalc/logarithms.htm</a>


lot's of stuff on the web.
it just takes a little looking to get the one that makes the most sense to you.