# Given $\log 6$ and $\log 8$,then the only logarithm that cannot be obtained without using the table is

A.$\log 64$

B.$\log 21$

C.$\log \dfrac{8}{3}$

D.$\log 9$

Last updated date: 24th Mar 2023

•

Total views: 306.9k

•

Views today: 3.84k

Answer

Verified

306.9k+ views

Hint: We need to evaluate each part using logarithmic properties and the given log values.

Given the values of $\log 6$ and $\log 8$. We need to find the values of another logarithm using them.

Let us assume, $y = \log 6$

We know that,$\log \left( {a \times b} \right) = \log a + \log b$.

Using this in above, we get

$y = \log \left( {2 \times 3} \right) = \log 2 + \log 3$ ………………...(1)

We also assume, $x = \log 8$

We know that,$\log \left( {{a^b}} \right) = b\log a$

Using this in above, we get

\[

x = \log \left( {{2^3}} \right) = 3\log 2 \\

\Rightarrow \dfrac{x}{3} = \log 2 \\

\] …………………….(2)

Form equations (1) and (2) , we get

\[

y = \log 2 + \log 3 = \dfrac{x}{3} + \log 3 \\

\Rightarrow \log 3 = y - \dfrac{x}{3} \\

\] ……………………..(3)

Now we try to evaluate each given part using the above obtained values in equations (2) and (3).

In part (A), we have

$\log 64$

Using $\log \left( {{a^b}} \right) = b\log a$ and equation (2) in above, we get

$\log {2^6} = 6\log 2 = 6 \times \dfrac{x}{3} = 2x$

Given we know the value of x, the above part can be calculated using given logarithms.

In part (B), we have

$\log 21$

Using $\log \left( {a \times b} \right) = \log a + \log b$in above, we get

$\log 21 = \log 7 \times 3 = \log 7 + \log 3$

In the above equation, we are not given the value of $\log 7$, hence we cannot find the value of this part without using the log table.

In part (C), we have

$\log \dfrac{8}{3}$

Using $\log \left( {\dfrac{a}{b}} \right) = \log a - \log b$ , $\log \left( {{a^b}} \right) = b\log a$, equation (2) and (3) in above, we get

$\log \dfrac{8}{3} = \log 8 - \log 3 = 3\log 2 - \log 3 = x - y + \dfrac{x}{3}$

Given we know the value of x and y, the above part can be calculated using given logarithms.

In part (D), we have

$\log 9$

Using $\log \left( {{a^b}} \right) = b\log a$ and equation (3) in above, we get

$\log {3^2} = 2\log 3 = 2y - \dfrac{{2x}}{3}$

Given we know the value of x and y, the above part can be calculated using given logarithms.

Hence only part (B) cannot be calculated without using the table in the above problem.

Note: The properties of logarithm should be kept in mind while solving the problems like above. The logarithm value can change with the change in the base. Hence it is crucial to convert base when necessary.

__Complete step-by-step answer:__Given the values of $\log 6$ and $\log 8$. We need to find the values of another logarithm using them.

Let us assume, $y = \log 6$

We know that,$\log \left( {a \times b} \right) = \log a + \log b$.

Using this in above, we get

$y = \log \left( {2 \times 3} \right) = \log 2 + \log 3$ ………………...(1)

We also assume, $x = \log 8$

We know that,$\log \left( {{a^b}} \right) = b\log a$

Using this in above, we get

\[

x = \log \left( {{2^3}} \right) = 3\log 2 \\

\Rightarrow \dfrac{x}{3} = \log 2 \\

\] …………………….(2)

Form equations (1) and (2) , we get

\[

y = \log 2 + \log 3 = \dfrac{x}{3} + \log 3 \\

\Rightarrow \log 3 = y - \dfrac{x}{3} \\

\] ……………………..(3)

Now we try to evaluate each given part using the above obtained values in equations (2) and (3).

In part (A), we have

$\log 64$

Using $\log \left( {{a^b}} \right) = b\log a$ and equation (2) in above, we get

$\log {2^6} = 6\log 2 = 6 \times \dfrac{x}{3} = 2x$

Given we know the value of x, the above part can be calculated using given logarithms.

In part (B), we have

$\log 21$

Using $\log \left( {a \times b} \right) = \log a + \log b$in above, we get

$\log 21 = \log 7 \times 3 = \log 7 + \log 3$

In the above equation, we are not given the value of $\log 7$, hence we cannot find the value of this part without using the log table.

In part (C), we have

$\log \dfrac{8}{3}$

Using $\log \left( {\dfrac{a}{b}} \right) = \log a - \log b$ , $\log \left( {{a^b}} \right) = b\log a$, equation (2) and (3) in above, we get

$\log \dfrac{8}{3} = \log 8 - \log 3 = 3\log 2 - \log 3 = x - y + \dfrac{x}{3}$

Given we know the value of x and y, the above part can be calculated using given logarithms.

In part (D), we have

$\log 9$

Using $\log \left( {{a^b}} \right) = b\log a$ and equation (3) in above, we get

$\log {3^2} = 2\log 3 = 2y - \dfrac{{2x}}{3}$

Given we know the value of x and y, the above part can be calculated using given logarithms.

Hence only part (B) cannot be calculated without using the table in the above problem.

Note: The properties of logarithm should be kept in mind while solving the problems like above. The logarithm value can change with the change in the base. Hence it is crucial to convert base when necessary.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE