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$

{\text{A}}{\text{. }}\dfrac{{4{\text{b}}}}{{{\text{c + a - b}}}} \\

{\text{B}}{\text{. }}\dfrac{{4{\text{c}}}}{{{\text{a + b - c}}}} \\

{\text{C}}{\text{. }}\dfrac{{4{\text{b}}}}{{{\text{c - a - b}}}} \\

{\text{D}}{\text{. }}\dfrac{{4{\text{a}}}}{{{\text{a + b - c}}}} \\

$

Answer

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Hint: Take log both sides of the equation ${6^{\text{x}}} = {7^{{\text{x + 4}}}}$ and use properties of logarithms.

Complete step-by-step answer:

Let,

log2 = a …………….(1)

log3 = b ……………..(2)

log7 = c ……………….(3)

${6^{\text{x}}} = {7^{{\text{x + 4}}}}$ ……………….(4)

As for any positive real number k, other than 1 such that ${{\text{k}}^{\text{m}}}{\text{ = x}}$ then , a logarithmic function can be defined as ${\text{m = lo}}{{\text{g}}_{\text{k}}}{\text{x}}$, where k is the base.

Now, in equation 4 we have, ${6^{\text{x}}}$ = ${7^{{\text{x + 4}}}}$ . On taking log both sides , we get

log(${6^{\text{x}}}$) = log(${7^{{\text{x + 4}}}}$)

Applying the property of logarithm which states ${\text{log(}}{{\text{a}}^{\text{n}}}) = {\text{nlog(a)}}$, we get

xlog6=(x+4)log7

$

{\text{xlog(3}} \times {\text{2) = xlog7 + 4log7}} \\

\\

$

Applying another property of logarithm which states ${\text{log(a}} \times {\text{b) = loga + logb}}$, we get

xlog3 + xlog2 – xlog7 = 4log7

Substituting the values of log2, log3 and log 7 from equation 1,2 and 3.

ax + bx – cx = 4c

x(a +b -c) = 4c

or, x = $\dfrac{{{\text{4c}}}}{{{\text{a + b - c}}}}$.

Answer is option (b).

Note: In these types of questions, the key concept is to remember the properties of logarithm. The logarithm question requires only two steps. Step 1 is to convert the equation into logarithmic form. Step 2, is apply the properties of logarithm and simplify it to the end.

Complete step-by-step answer:

Let,

log2 = a …………….(1)

log3 = b ……………..(2)

log7 = c ……………….(3)

${6^{\text{x}}} = {7^{{\text{x + 4}}}}$ ……………….(4)

As for any positive real number k, other than 1 such that ${{\text{k}}^{\text{m}}}{\text{ = x}}$ then , a logarithmic function can be defined as ${\text{m = lo}}{{\text{g}}_{\text{k}}}{\text{x}}$, where k is the base.

Now, in equation 4 we have, ${6^{\text{x}}}$ = ${7^{{\text{x + 4}}}}$ . On taking log both sides , we get

log(${6^{\text{x}}}$) = log(${7^{{\text{x + 4}}}}$)

Applying the property of logarithm which states ${\text{log(}}{{\text{a}}^{\text{n}}}) = {\text{nlog(a)}}$, we get

xlog6=(x+4)log7

$

{\text{xlog(3}} \times {\text{2) = xlog7 + 4log7}} \\

\\

$

Applying another property of logarithm which states ${\text{log(a}} \times {\text{b) = loga + logb}}$, we get

xlog3 + xlog2 – xlog7 = 4log7

Substituting the values of log2, log3 and log 7 from equation 1,2 and 3.

ax + bx – cx = 4c

x(a +b -c) = 4c

or, x = $\dfrac{{{\text{4c}}}}{{{\text{a + b - c}}}}$.

Answer is option (b).

Note: In these types of questions, the key concept is to remember the properties of logarithm. The logarithm question requires only two steps. Step 1 is to convert the equation into logarithmic form. Step 2, is apply the properties of logarithm and simplify it to the end.

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