Question

if a=b^2=c^3=d^4 then the value of log(abcd) to base a

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{a=b^2=c^3=d^4}$

$\underline{\textbf{To find:}}$

$\textsf{The value of}$

$\mathsf{log\,_a\,(abcd)}$

$\underline{\textbf{Solution:}}$

$\mathsf{a=b^2=c^3=d^4=k\;\;(say)}$

$\mathsf{a=k}$

$\mathsf{b^2=k\;\implies\;b=k^\dfrac{1}{2}}$

$\mathsf{c^3=k\;\implies\;c=k^\dfrac{1}{3}}$

$\mathsf{d^4=k\;\implies\;d=k^\dfrac{1}{4}}$

$\mathsf{Now,}$

$\mathsf{log\,_a\,(abcd)}$

$\mathsf{=log\,_k\left(k\,k^{\dfrac{1}{2}}\,k^{\dfrac{1}{3}}\,k^{\dfrac{1}{4}}\right)}$

$\mathsf{=log\,_k\left(k^\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)\right)}$

$\mathsf{=log\,_k\left(k^\left(\dfrac{12+6+4+3}{12}\right)\right)}$

$\mathsf{=log\,_k\,\left(k^\left(\dfrac{25}{12}\right)\right)}$

$\textsf{Using power rule of logarithm, we get}$

$\mathsf{=\dfrac{25}{12}\;log\,_k\,k}$

$\mathsf{=\dfrac{25}{12}\;(1)}\;\;\mathsf{(\because\;log\,_aa=1)}$

$\mathsf{=\dfrac{25}{12}}$

$\implies\boxed{\mathsf{log\,_a\,(abcd)=\dfrac{25}{12}}}$

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Answer 2

Answer:

Concept:

A base must be raised to a certain exponent or power, or logarithm, in order to produce a specific number. If  $b^{x}$ = n  then x is the logarithm of n to the base b, which is expressed mathematically as x = $log_{b}$ n.

Step-by-step explanation:

Given:

$Given \ a=b^{2}=c^{3}=d^{4}$

Find:

$what \ is \ the \ value \ of \log _{a} a b c d= ?$
Solution:
        $\begin{aligned}\log _{a}^{a b c d} &=\log _{a} a\left(b^{2}\right)^{1 / 2}\left(c^{3)^{1 / 3}}\left(d^{4}\right)^{\frac{1}{4}}\right.\\&=\log _{a}+\frac{1}{2} \log _{a}^{b^{2} +\frac{1}{3} \log c^{3}+1 / \log _{a}^{4}} \\ &=1+\frac{1}{2} \log a+\frac{1}{3} \log a+\frac{1}{4} \log a \\&=1+\frac{1}{2} \cdot 1+\frac{1}{3} \cdot 1+\frac{1}{4} \cdot 1 \\&=\frac{12+6+4+3}{12} \\\log _{a} a b c d &=\frac{25}{12}\end{aligned}$

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