Prove that xloga to base b = a log x to base b
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Answer:
Answer and explanation:
To prove : \log_{ab}x=\frac{(\log_a x)(\log_b x)}{\log_a x+\log_b x}log
ab
x=
log
a
x+log
b
x
(log
a
x)(log
b
x)
Proof :
Taking RHS,
\frac{(\log_a x)(\log_b x)}{\log_a x+\log_b x}
log
a
x+log
b
x
(log
a
x)(log
b
x)
=\frac{(\log_a x)(\log_b x)}{\frac{1}{\log_x a}+\frac{1}{\log_x b}}=
log
x
a
1
+
log
x
b
1
(log
a
x)(log
b
x)
=\frac{(\log_a x)(\log_b x)}{\frac{\log_x a+\log_x b}{(\log_x a)(\log_x b)}}=
(log
x
a)(log
x
b)
log
x
a+log
x
b
(log
a
x)(log
b
x)
=\frac{(\log_a x)(\log_b x)(\log_x a)(\log_x b)}{\log_x a+\log_x b}=
log
x
a+log
x
b
(log
a
x)(log
b
x)(log
x
a)(log
x
b)
=\frac{1}{\log_x a+\log_x b}=
log
x
a+log
x
b
1
=\frac{1}{\log_x (ab)}=
log
x
(ab)
1
=\log_{ab}x=log
ab
x
= LHS
Hence proved.
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