Question

pls solve it
and also tell that can we do it like this
log 5 - log 10​

Answer 1

We know it is a number such that

$\bold{10^{log_{10}\frac{5}{10} }=\dfrac{5}{10} }$

It is possible to decompose the right hand side

$\bold{10^{log_{10}\frac{5}{10} }=5\times\dfrac{1}{10} }$

Negative power rule applies here

$\bold{10^{log_{10} \frac{5}{10} }=5\times 10^{-1}}$

According to the right hand side, we change the left hand side

$\bold{10^{log_{10}5 }\times10^{log_{10}10^{-1}}=5\times 10^{-1}}$

Now power rule can simplify the power

$\bold{10^{log_{10}5 +log_{10}10^{-1}}=5\times \dfrac{1}{10} }$

We take logarithm base 10 on both sides

$\bold{log_{10}5+ log_{10}10^{-1} =log_{10}\dfrac{5}{10} }$

Take the negative power, hence shown

$\bold{\therefore log_{10}5- log_{10}10 =log_{10}\dfrac{5}{10} }$

Or we can decompose once more

$\bold{\cancel{log_{10}5}- log_{10}2- \cancel{log_{10}5} =log_{10}\dfrac{5}{10} }$

$\bold{\therefore -log_{10}2=log_{10}\dfrac{5}{10} }$

Extra information:

We are able to take logarithm on both sides as a logarithm function and an exponential function is the inverse function of each other.

Let $\bold{f(x)=10^x}$

Then $\bold{f^{-1}(x)=log_{10}x}$

We know $\bold{f(f^{-1}(x))=x}$.

So we get the power when logarithm is taken on exponents.

Answer 2

Answer:

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