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# Find the value of x, if log2 = a, log3 = b , log7 = c and ${6^{\text{x}}} = {7^{{\text{x + 4}}}}$ ${\text{A}}{\text{. }}\dfrac{{4{\text{b}}}}{{{\text{c + a - b}}}} \\ {\text{B}}{\text{. }}\dfrac{{4{\text{c}}}}{{{\text{a + b - c}}}} \\ {\text{C}}{\text{. }}\dfrac{{4{\text{b}}}}{{{\text{c - a - b}}}} \\ {\text{D}}{\text{. }}\dfrac{{4{\text{a}}}}{{{\text{a + b - c}}}} \\$

Last updated date: 29th Mar 2023
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Hint: Take log both sides of the equation ${6^{\text{x}}} = {7^{{\text{x + 4}}}}$ and use properties of logarithms.

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