Question

9 If a = 1 + log x (yz), b = 1 + log y (zx) and c = 1 + log z (xy) , then show that ab + bc + ca = abc.​

Answer 1

Answer:

See the explanation

Step-by-step explanation:

Given,

   $a=1+\log_x(yz),b=1+\log_y(xz),c=1+\log_z(xy)$

 $\Rightarrow a=\log_x(x)+\log_x(yz),b=\log_y(y)+\log_y(xz),c=\log_z(z)+\log_z(xy)$

                                   $\boxed{\because \log_aa=1}$

$\Rightarrow a=\log_x(xyz),b=\log_y(xyz),c=\log_z(xyz)$

                             $\boxed{\because \log_ax+\log_ay=\log_z(xy)}$

$\Rightarrow \dfrac{1}{a}=\dfrac{1}{\log_x(xyz)}=\log_{xyz}(x),\dfrac{1}{b}=\dfrac{1}{\log_y(xyz)}=\log_{xyz}(y),\dfrac{1}{c}=\dfrac{1}{\log_z(xyz)}=\log_{xyz}(z)$

                             $\boxed{\because \log_ab=\dfrac{1}{\log_ba}}$

$\Rightarrow \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\log_{xyz}(x)+\log_{xyz}(y)+\log_{xyz}(z)=\log_{xyz}(xyz)=1\\\\\Rightarrow \dfrac{bc+ca+ab}{abc}=1\\\\\Rightarrow ab+bc+ca=abc$

Hence proved

Answer 2

a=1+\log_x(yz),b=1+\log_y(xz),c=1+\log_z(xy)a=1+log

x

(yz),b=1+log

y

(xz),c=1+log

z

(xy)

\Rightarrow a=\log_x(x)+\log_x(yz),b=\log_y(y)+\log_y(xz),c=\log_z(z)+\log_z(xy)⇒a=log

x

(x)+log

x

(yz),b=log

y

(y)+log

y

(xz),c=log

z

(z)+log

z

(xy)

\boxed{\because \log_aa=1}

∵log

a

a=1

\Rightarrow a=\log_x(xyz),b=\log_y(xyz),c=\log_z(xyz)⇒a=log

x

(xyz),b=log

y

(xyz),c=log

z

(xyz)

\boxed{\because \log_ax+\log_ay=\log_z(xy)}

∵log

a

x+log

a

y=log

z

(xy)

\Rightarrow \dfrac{1}{a}=\dfrac{1}{\log_x(xyz)}=\log_{xyz}(x),\dfrac{1}{b}=\dfrac{1}{\log_y(xyz)}=\log_{xyz}(y),\dfrac{1}{c}=\dfrac{1}{\log_z(xyz)}=\log_{xyz}(z)⇒

a

1

=

log

x

(xyz)

1

=log

xyz

(x),

b

1

=

log

y

(xyz)

1

=log

xyz

(y),

c

1

=

log

z

(xyz)

1

=log

xyz

(z)

\boxed{\because \log_ab=\dfrac{1}{\log_ba}}

∵log

a

b=

log

b

a

1

\begin{gathered}\Rightarrow \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\log_{xyz}(x)+\log_{xyz}(y)+\log_{xyz}(z)=\log_{xyz}(xyz)=1\\\\\Rightarrow \dfrac{bc+ca+ab}{abc}=1\\\\\Rightarrow ab+bc+ca=abc\end{gathered}

⇒

a

1

+

b

1

+

c

1

=log

xyz

(x)+log

xyz

(y)+log

xyz

(z)=log

xyz

(xyz)=1

⇒

abc

bc+ca+ab

=1

⇒ab+bc+ca=abc

Hence proved