ha lagta hai ban ho gaya
Answers
Explanation:
Let's first recall Identities Used to solve this problem!!!
\begin{gathered}\boxed{ \sf{ \: log_{x}(x) = 1 \: }} \\ \\ \end{gathered}
log
x
(x)=1
\begin{gathered}\boxed{ \sf{ \: log_{x}(a) + log_{a}(b) = log_{x}(ab) \: }} \\ \\ \end{gathered}
log
x
(a)+log
a
(b)=log
x
(ab)
(a)−log
a
(b)=log
x
b
a
\begin{gathered}\boxed{ \sf{ \: log_{a}(b) \: = \: \frac{1}{ log_{b}(a) } \: \: }} \\ \\ \\ \end{gathered}
log
a
(b)=
log
b
(a)
1
Let's solve the problem now!!!
\large\underline{\sf{Solution-}}
Solution−
Given that,
\begin{gathered}\sf \: x = 1 + log_{a}(bc) \\ \\ \end{gathered}
x=1+log
a
(bc)
can be rewritten as
\begin{gathered}\sf \: x = 1 + log_{a}\bigg(\dfrac{abc}{a} \bigg) \\ \\ \end{gathered}
x=1+log
a
(
a
abc
)
\begin{gathered}\sf \: x = 1 + log_{a}(abc) - log_{a}(a) \\ \\ \end{gathered}
x=1+log
a
(abc)−log
a
(a)
\begin{gathered}\sf \: x = 1 + log_{a}(abc) - 1 \\ \\ \end{gathered}
x=1+log
a
(abc)−1
\begin{gathered}\sf \: x = log_{a}(abc) \\ \\ \end{gathered}
x=log
a
(abc)
can be further rewritten as
\begin{gathered}\sf \: x = \dfrac{1}{ log_{abc}(a) } \\ \\ \end{gathered}
x=
log
abc
(a)
1
\begin{gathered}\bf\implies \:\dfrac{1}{x} = log_{abc}(a) - - - (1) \\ \\ \end{gathered}
⟹
x
1
=log
abc
(a)−−−(1)
Similarly,
\begin{gathered}\bf\implies \:\dfrac{1}{y} = log_{abc}(b) - - - (2) \\ \\ \end{gathered}
⟹
y
1
=log
abc
(b)−−−(2)
and Similarly
\begin{gathered}\bf\implies \:\dfrac{1}{z} = log_{abc}(c) - - - (3) \\ \\ \end{gathered}
⟹
z
1
=log
abc
(c)−−−(3)
On adding equation (1), (2) and (3), we get
\begin{gathered}\sf \: \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \\ \\ \end{gathered}
x
1
+
y
1
+
z
1
\begin{gathered}\sf \: = \: log_{abc}(a) \: + \: log_{abc}(b) \: + \: log_{abc}(c) \\ \\ \end{gathered}
=log
abc
(a)+log
abc
(b)+log
abc
(c)
\begin{gathered}\sf \: = \: log_{abc}(abc) \\ \\ \end{gathered}
=log
abc
(abc)
\begin{gathered}\sf \: = \: 1 \\ \\ \end{gathered}
=1
\begin{gathered}\bf\implies \:\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1 \\ \\ \end{gathered}
⟹
x
1
+
y
1
+
z
1
=1
\begin{gathered}\bf\implies \:\dfrac{yz + zx + xy}{xyz} = 1 \\ \\ \end{gathered}
⟹
xyz
yz+zx+xy
=1
\begin{gathered}\bf\implies \:xy + yz + zx = xyz \\ \\ \end{gathered}
⟹xy+yz+zx=xyz
\rule{190pt}{2pt}
{{ \mathfrak{Additional\:Information}}}AdditionalInformation
\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ log_{x}(x) = 1}\\ \\ \bigstar \: \bf{ log_{x}( {x}^{y} ) = y}\\ \\ \bigstar \: \bf{ log_{ {x}^{z} }( {x}^{w} ) = \dfrac{w}{z} }\\ \\ \bigstar \: \bf{ log_{a}(b) = \dfrac{logb}{loga} }\\ \\ \bigstar \: \bf{ {e}^{logx} = x}\\ \\ \bigstar \: \bf{ {e}^{ylogx} = {x}^{y}}\\ \\ \bigstar \: \bf{log1 = 0}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}
MoreFormulae
MoreFormulae
★log
x
(x)=1
★log
x
(x
y
)=y
★log
x
z
(x
w
)=
z
w
★log
a
(b)=
loga
logb
★e
logx
=x
★e
ylogx
=x
y
★log1=0