Question

the first amd best answer will be marked as brainliest by me....promise
Utkarsh Pratham
according class 9​

Answer 1

◘ Given ◘

$\bullet\ \sf{(xy)^{(a-1)}=z}$  

$\bullet\ \sf{(yz)^{(b-1)}=x}$  

$\bullet\ \sf{(xz)^{(c-1)}=y}$  

$\bullet\ \sf{xyz \neq -1,\ 0\ or\ 1}$

◘ To Find ◘

The value which is equal to ab + bc + ca.

◘ Solution ◘

$\sf{(xy)^{(a-1)}=z}\\\\\sf{\implies log\ xy^{(a-1)}=log\ z}$

(applying log to both the sides)

$\sf{\implies (a-1)log\ xy=log\ z}$

(As log m^n = n*log m)

$\sf{\implies \dfrac{(a-1)log\ xy}{log\ z}=1}\\\\\sf{\implies \dfrac{log\ xy}{log\ z}=\dfrac{1}{a-1} }\\\\\sf{\implies a-1=\dfrac{log\ z}{log\ xy} }\\\\\sf{\implies a=\dfrac{log\ z}{log\ xy}+1}\\\\\sf{\implies a=\dfrac{log\ z+log\ xy}{log\ xy} }\\\\\sf{\implies a= \dfrac{log\ xyz}{log\ xy} }$

(As log ab = log a + log b)

Similarly,

$\bigstar\ \sf{b= \dfrac{log\ xyz}{log\ zy} }$

$\bigstar\ \sf{c= \dfrac{log\ xyz}{log\ zx} }$

$\rule{160}4$

Now,

$\sf{a+b+c=\dfrac{log\ xyz}{log\ xy} +\dfrac{log\ xyz}{log\ zy} +\dfrac{log\ xyz}{log\ zx} }$

$\sf{\implies \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{log\ xy}{log\ xyz} +\dfrac{log\ yz}{log\ xyz} +\dfrac{log\ xz}{log\ xyz} }$

$\sf{\implies \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{log\ xy+log\ yz+log\ xz}{log\ xyz} }$

(As log ab = log a + log b)

$\sf{\implies \dfrac{bc+ac+ab}{abc}=\dfrac{log\ ( xy\times yz\times xz)}{log\ xyz} }$

$\sf{\implies \dfrac{ab+bc+ca}{abc}=\dfrac{log\ (x^2 \times y^2 \times z^2)}{log\ xyz} }$

$\sf{\implies \dfrac{ab+bc+ca}{abc}=\dfrac{log\ (xyz)^2}{log\ xyz} }$

$\sf{\implies \dfrac{ab+bc+ca}{abc}= \dfrac{2 \times log\ xyz}{log\ xyz} }$

(As log m^n = n*log m)

$\sf{\implies \dfrac{ab+bc+ca}{abc}= 2 }$

$\large \boxed{\sf{\implies ab+bc+ca= 2abc }}$

$\rule{160}4$

Hope this helps -,-

Answer 2

$\huge\sf\pink{Answer}$

☞ ab+bc+ca = 2abc

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ $\sf{(xy)^{(a-1)}=z}$

✭ $\sf{(yz)^{(b-1)}=x}$

✭ $\sf{(xz)^{(c-1)}=y}$

✭ $\sf{xyz \ne -1,\ 0\ or\ 1}$

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ Which of the above value is equal to ab+bc+ca

$\rule{110}1$

$\huge\sf\purple{Steps}$

$\bullet\:\underline{\textsf{As Per the Question}}$

➢ $\sf{(xy)^{(a-1)}=z}$

➢ $\sf{log\ xy^{(a-1)}=log\ z}$

Here we shall apply log on both sides,

➝ $\sf{(a-1)log\ xy=log\ z}$

$\bigg\lgroup \sf log \ m^n=n\times log \ m \bigg\rgroup$

➝ $\sf{\dfrac{(a-1)log\ xy}{log\ z}=1}$

➝ $\sf{\dfrac{log\ xy}{log\ z}=\dfrac{1}{a-1}}$

➝ $\sf{a-1=\dfrac{log\ z}{log\ xy}}$

➝ $\sf{a=\dfrac{log\ z}{log\ xy}+1}$

➝ $\sf{a=\dfrac{log\ z+log\ xy}{log\ xy}}$

➝ $\sf{a= \dfrac{log\ xyz}{log\ xy}}$

$\bigg\lgroup \sf log \ ab = log \ a + log \ b\bigg\rgroup$

$\bigg\lgroup\sf{b= \dfrac{log\ xyz}{log\ zy}}\bigg\rgroup$

$\bigg\lgroup\sf{c= \dfrac{log\ xyz}{log\ zx}}\bigg\rgroup$

So now,

$\small\underline{\boxed{\sf{a+b+c=\dfrac{log\ xyz}{log\ xy} +\dfrac{log\ xyz}{log\ zy} +\dfrac{log\ xyz}{log\ zx}}}}$

➳ $\sf{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{log\ xy}{log\ xyz} +\dfrac{log\ yz}{log\ xyz} +\dfrac{log\ xz}{log\ xyz}}$

➳ $\sf{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{log\ xy+log\ yz+log\ xz}{log\ xyz}}$

➳ $\sf{\dfrac{bc+ac+ab}{abc}=\dfrac{log\ ( xy\times yz\times xz)}{log\ xyz}}$

➳ $\sf{\dfrac{ab+bc+ca}{abc}=\dfrac{log\ (x^2 \times y^2 \times z^2)}{log\ xyz}}$

➳ $\sf{\dfrac{ab+bc+ca}{abc}=\dfrac{log\ (xyz)^2}{log\ xyz}}$

➳ $\sf{\dfrac{ab+bc+ca}{abc}= \dfrac{2 \times log\ xyz}{log\ xyz}}$

➳ $\sf{\dfrac{ab+bc+ca}{abc}= 2}$

➳ $\sf\orange{ab+bc+ca=2abc}$

$\rule{170}3$