Question

Log5 2.log11 5.log32 11 =

Answer 1

Step-by-step explanation:

a

log b = log a/logb

= log2/log5 × log5/log11 × log 11/log32

cancel all the same logs & then;

5

=log2/log2 = log2/5 log2

then log2/log 2 cancelled.. remaining

=1/5

formula :-

b

log a = b loga

Answer 2

$\displaystyle \sf{ log_{5} 2 \: . \: log_{11} 5 \: . \:log_{32} 11 } = \frac{1}{5}$

Given :

$\displaystyle \sf{ log_{5} 2 \: . \: log_{11} 5 \: . \:log_{32} 11 }$

To find :

To simplify the expression

Formula :

$\displaystyle \sf{ log_{a} b = \frac{log \: b}{log \: a} }$

Solution :

Step 1 of 2 :

Write down the given expression

The given expression is

$\displaystyle \sf{ log_{5} 2 \: . \: log_{11} 5 \: . \:log_{32} 11 }$

Step 2 of 2 :

Simplify the given expression

$\displaystyle \sf{ log_{5} 2 \: . \: log_{11} 5 \: . \:log_{32} 11 }$

$\displaystyle \sf{ = \frac{log \: 2}{log \: 5} \times \frac{log \: 5}{log \: 11} \times \frac{log \: 11}{log \: 32}}$

$\displaystyle \sf{ = \frac{log \: 2}{log \: 32} }$

$\displaystyle \sf{ = \frac{log \: 2}{log \: {2}^{5} } }$

$\displaystyle \sf{ = \frac{log \: 2}{5log \: {2}^{} } }$

$\displaystyle \sf{ = \frac{1}{5} }$

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