Question

a powerx=b powery= c powerz=d powerw then log(abc) base d=​

Answer 1

Answer:

$log_d{abc}=\frac{yzw+wxz+wxy}{xyz}$

Step-by-step explanation:

Formula used:

$1.\:log_a{b}=\frac{1}{log_b{a}}$

$2.\:log_aM^n=n\:log_aM$

Given:

$a^x=b^y=c^z=d^w=k(say)$

Then,

$a^x=k\:\implies\:a=k^{\frac{1}{x}}$

$b^y=k\:\implies\:b=k^{\frac{1}{y}}$

$c^z=k\:\implies\:c=k^{\frac{1}{z}}$

$d^w=k\:\implies\:d=k^{\frac{1}{w}}$

Now,

$log_d{abc}$

$=log_d{k^{\frac{1}{x}}.k^{\frac{1}{y}}.k^{\frac{1}{z}}}$

$=log_d{k^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$

$={\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}log_d{k}$

$=(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{log_k{d}})$

$=(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{log_k{k^{\frac{1}{w}}}})$

$=(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{\frac{1}{w}log_kk})$

$=(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{w}{log_kk})$

$=(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{w}{1})$

$=w(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$

$=w(\frac{yz+xz+xy}{xyz})$

$=\frac{yzw+wxz+wxy}{xyz}$