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Pavel [41]
2 years ago
7

Prove that :

Mathematics
2 answers:
tia_tia [17]2 years ago
6 0

\bigstar \: {\large{\textsf{\textbf{\underline{\underline{Concept :}}}}}}

• Some of the properties of the log are as follows :

1) <u>Product law </u><u>-</u>

\star \: \sf log(mnp) = \underline{{\boxed{{\sf logm +logn + logp}}}}

2) <u>Divisio</u><u>n</u><u> </u><u>law</u><u> </u><u>-</u>

\star \: \sf log( \dfrac{m}{n} ) = \underline{{\boxed{{\sf logm  - logn }}}}

3) <u>Power law -</u>

\star \: \sf log \:  {m}^{n} = \underline{{\boxed{{\sf n \: log m }}}}

4) We should also know that -

\star \: \sf log 10 = \underline{{\boxed{{\sf 1 }}}}

• To prove R.H.S equals to L.H.S, there exists three conditions :

1) To make R.H.S equals to L.H.S

2) To make L.H.S equals to R.H.S

3) To simplify both side equations and make them equal to a single value

In proving these questions, we're going to apply 2nd condition though all the three conditions described above are convertible to each other.

<u>Now, let's </u><u>start</u><u>!</u>

\: {\large{\textsf{\textbf{\underline{\underline{Solution:}}}}}}

\sf \red{ Question\: 1}

<u>Taking</u><u> </u><u>L.H.S</u>

\sf log630

✦ Prime factors of 630 -

\begin{gathered}\begin{gathered}{\begin{array}{ c|c}2&630 \\ \hline 3&315 \\ \hline 3&105\\ \hline5&35 \\ \hline 7&7\\  \hline &1\end{array}}\end{gathered}\end{gathered}

\implies \sf log(2 \times  {3}^{2}  \times 5 \times 7)

• Using first property

\implies \sf log2  + \underline{log  {3}^{2} }+   log5  + log 7

• Using <u>third</u> property

\implies \sf log2  +2 \: log3 +   log5  + log 7

\therefore \: \sf log630 =  \underline{{\boxed{ \red{{\sf log2+2 \: log3 + log5 + log7}}}}}

\sf \green{ Question\: 2}

<u>Taking L.H.S</u>

\sf log10 + log100 + log1000 + log10000

\implies \sf log10 + log {10}^{2}  + log {10}^{3} + log {10}^{4}

• Using third property

\implies \sf log10 +2 \:  log 10  + 3 \: log 10 + 4 \: log 10

• As we know log10 = 1

\implies \sf 1+2  (1) + 3 (1)+ 4 (1)

\implies \sf 1+2  + 3 + 4

\implies \sf  \green{10}

\therefore \: \sf   log10 + log100 + log1000 + log10000=  \underline{{\boxed{ \green{{\sf 10}}}}}

\rule{280pt}{2pt}

alisha [4.7K]2 years ago
6 0

Number 1

log 630

=log (2×3×3×5×7)

=log2+log 3^2-log 5+log7

= log 2+2 log 3 + log 5 + log 7

Number 2

log 10+ log 100+ log 1000+ log 10000

= log 10+2 log 10+ 3 log 10+ 4 log 10

= 10 log 10

= 10

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