Question

evaluate 2log3 + 3log4 + 4log5 - 2log6​

Answer 1

We have:

2$\log 3$ + 3$\log 4$ + 4$\log 5$ - 2$\log 6$

We have to find, the value of 2$\log 3$ + 3$\log 4$ + 4$\log 5$ - 2$\log 6$ = ?

Solution:

∴ 2$\log 3$ + 3$\log 4$ + 4$\log 5$ - 2$\log 6$

= 2$\log 3$ + 3$\log 2^2$ + 4$\log 5$ - 2$\log (2\times 3)$

Using the logarithm identity:

$\log m^n=n\log m$

and  $\log (m\times n)=\log m + \log n$

= 2$\log 3$ + (3 × 2)$\log 2$ + 4$\log 5$ - 2$\log 2$ - 2$\log 3$

= (2$\log 3$ - 2$\log 3$ )+ (6$\log 2$ - 2$\log 2$) + 4$\log 5$

= 0 + 4$\log 2$ + 4$\log 5$

= 4$\log 2$ + 4$\log 5$

= 4($\log 2$ + $\log 5$)

= 4$\log (2\times 5)$  [∵ $\log (m\times n)=\log m + \log n$]

= 4$\log (2\times 5)$

= 4 × 1 [∵ $\log 10$ = 1]

= 4

∴ 2$\log 3$ + 3$\log 4$ + 4$\log 5$ - 2$\log 6$ = 4

Thus, the value of 2$\log 3$ + 3$\log 4$ + 4$\log 5$ - 2$\log 6$ is "equal to 4".