Question

Log16^*+log4^*+log2^*=7​

Answer 1

`x = 16`

Explanation:

Given

`log_(16)(x) + log_4(x) + log_2(x) = 7`

Required

Solve for x

`log_(16)(x) + log_4(x) + log_2(x) = 7`

Change base of 16 and base of 4 to base 2

`(log_2(x))/(log_2(16)) + (log_2(x))/(log_2(4)) + log_2(x) = 7`

Express 16 and 4 as 2^4 and 2^2 respectively

`(log_2(x))/(log_2(2^4)) + (log_2(x))/(log_2(2^2)) + log_2(x) = 7`

The above can be rewritten as:

`(log_2(x))/(4log_22) + (log_2(x))/(2log_22) + log_2(x) = 7`

`log_22 = 1`

So, we have:

`(log_2(x))/(4*1) + (log_2(x))/(2*1) + log_2(x) = 7`

`(1)/(4)log_2(x) + (1)/(2)log_2(x) + log_2(x) = 7`

Multiply through by 4

`4((1)/(4)log_2(x) + (1)/(2)log_2(x) + log_2(x)) = 7 * 4`

`log_2(x) + 2}log_2(x) + 4log_2(x) = 28`

`7log_2(x) = 28`

Divide through by 7

`(7log_2(x))/(7) = (28)/(7)`

`log_2(x) = 4`

Apply the following law of logarithm:

If
`log_ab = c` Then
`b = a^c`

So, we have:

`x = 2^4`

`x = 16`