# 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.$\log 9$

Answer

Verified

359.7k+ 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.

Last updated date: 16th Sep 2023

•

Total views: 359.7k

•

Views today: 8.59k