Question

Use the following steps to prove that log b(xy)- log bx+ log by.

Answer 1

Answer with Step-by-step explanation:

a.$x=b^p$

$y=b^q$

Taking both sides log

$log x=plog b$

Using identity:$logx^y=ylogx$

$p=\frac{logx}{log b}=log_b x$

Using identity:$log_x y=\frac{log y}{log x}$

$log y=qlog b$

$q=\frac{log y}{log b}=log_b y$

b.$xy=b^pb^q$

We know that

$x^a\cdot x^b=x^{a+b}$

Using identity

$xy=b^{p+q}$

c.$log_b(xy)=log_b(b^{p+q})$

$log_b(xy)=(p+q)log_b b$

Substitute the values then we get

$log_b(xy)=(log_b x+log_b y)$

By using $log_b b=1$

Hence, $log_b(xy)=log_b x+log_b y$