# Thread: Right angle triangle - logarithm problem

1. ## Right angle triangle - logarithm problem

Prove that if a and b are the lengths of the legs and c is the length of hypotenuse of a right angle triangle, c-b not equal to 1; c+b not equal to 1 then $log_{c+b}a+log_{c-b}a=2log_{c+b}alog_{c-b}a$

2. ## Re: Right angle triangle - logarithm problem

Originally Posted by sachinrajsharma
Prove that if a and b are the lengths of the legs and c is the length of hypotenuse of a right angle triangle, c-b not equal to 1; c+b not equal to 1 then $log_{c+b}a+log_{c-b}a=2log_{c+b}alog_{c-b}a$
1. Use the base-change-formula: $\log_b(a)=\frac{\ln(a)}{\ln(b)}$

$log_{c+b}a+log_{c-b}a=2log_{c+b}alog_{c-b}a~\implies~ \\ \\ \frac{\ln(a)}{\ln(c+b)}+\frac{\ln(a)}{\ln(c-b)}= 2 \cdot \frac{\ln(a)}{\ln(c+b)} \cdot \frac{\ln(a)}{\ln(c-b)}$

3. Divide through by ln(a) and multiply through by $\ln(c+b) \cdot \ln(c-b)$ . You'll get:

$\ln(c-b) + \ln(c+b) = 2\ln(a)$

Use laws of logarithms. You'll get:

$(c-b)(c+b)=a^2$

Expand and you'll get the Pythagorean theorem.