Question

Prove that log (51/80)+log(44/85)-log(99/160)-log(8/15)=0

Answer 1

refer to the image for proof

Answer 2

Answer:-

0

Step-by-step explanation:

We have to prove

$\mathbf{log(\frac{51}{80})\ +\ log(\frac{44}{85})\ -\ log(\frac{99}{160})\ -\ log(\frac{8}{15})\ =\ 0}$

Taking L.H.S

$log(\frac{51}{80})\ +\ log(\frac{44}{85})\ -\ log(\frac{99}{160})\ -\ log(\frac{8}{15})$

$log(\frac{51}{80})\ +\ log(\frac{44}{85})\ -\ [log(\frac{99}{160})\ +\ log(\frac{8}{15})]$

Using identity of log i.e log(A) + log(B) = log(A×B)

$log(\frac{51}{80}\ \times \frac{44}{85})\ -\ [log(\frac{99}{160}\ \times\ \frac{8}{15})]$

Again using the Identity of Log i.e $Log(\frac{A}{B})\ =\ Log(A) - Log(B)$

$Log(\frac{51\times 44\times 160\times 15}{80\times 85\times 99\times 8})$

On solving we get,

Log(1) = 0

This is equal to R.H.S

Hence Proved