Question

If $\log 2=a,\log 3=b,\log 7=c$ and $6^x=7^{x+4},$ then $x$ is equal to

• A. $\frac{4b}{c+a-b}$
• B. $\frac{4c}{a+b-c}$
• C. $\frac{4b}{c-a-b}$
• D. $\frac{4a}{a+b-c}$

Answer 1

Answer: (B)

$\frac{4c}{a+b-c}$

$log2=a$ $log3=b$ $log7=c.$

$6^x=7^{x+4}$

$(\frac{6}{7}{)}^x=7^4$

Taking logarithms,

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

$x[log6-log7]=4log7$

$x=\frac{4log7}{log6-log7}$

$x=\frac{4log7}{log2+log3-log7}$

$x=\frac{4c}{a+b-c}$