Question

expand log 1125/288??????​

Answer 1

Answer:

$\large\bold\red{0.57}$

Step-by-step explanation:

Given,

$log( \frac{1125}{288} ) \\ \\ = log( \frac{125 \times 9}{32 \times 9} ) \\ \\ = log( \frac{125}{32} )$

Now,

we know that,

$\large \bold{log( \frac{x}{y} ) = log(x) - log(y) }$

Therefore,

we get,

$= log(125) - log(32)$

$= log( {5}^{3} ) - log( {2}^{5} )$

But,

we know that,

$\large\bold{log( {x}^{y} ) = y log(x)}$

Therefore,

we get,

$= 3 log(5) - 5 log(2)$

But,

$\bold{ log(5) = 0.69 \: \: \: and \: \: \: log(2) = 0.30}$

Thus,

pUtting the values,

we get,

$= (3 \times 0.69) - (5 \times 0.30) \\ \\ = 2.07 - 1.50 \\ \\ = 0.57$

Hence,

0.57 is required answer.

Answer 2

Given,

\begin{gathered} log( \frac{1125}{288} ) \\ \\ = log( \frac{125 \times 9}{32 \times 9} ) \\ \\ = log( \frac{125}{32} ) \end{gathered}log(2881125)=log(32×9125×9)=log(32125)

Now,

we know that,

\large \bold{log( \frac{x}{y} ) = log(x) - log(y) }log(yx)=log(x)−log(y)

Therefore,

we get,

= log(125) - log(32)=log(125)−log(32)

= log( {5}^{3} ) - log( {2}^{5} )=log(53)−log(25)

But,

we know that,

\large\bold{log( {x}^{y} ) = y log(x)}log(xy)=ylog(x)

Therefore,

we get,

= 3 log(5) - 5 log(2)=3log(5)−5log(2)

But,

\bold{ log(5) = 0.69 \: \: \: and \: \: \: log(2) = 0.30}log(5)=0.69andlog(2)=0.30

Thus,

pUtting the values,

we get,

\begin{gathered} = (3 \times 0.69) - (5 \times 0.30) \\ \\ = 2.07 - 1.50 \\ \\ = 0.57\end{gathered}=(3×0.69)−(5×0.30)=2.07−1.50=0.57

Hence,

0.57 is required answer.