Question

Log16^*+log4^*+log2^*=7​

Answer 1

Answer:

$x = 16$

Step-by-step 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

$\frac{log_2(x)}{log_2(16)} + \frac{log_2(x)}{log_2(4)} + log_2(x) = 7$

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

$\frac{log_2(x)}{log_2(2^4)} + \frac{log_2(x)}{log_2(2^2)} + log_2(x) = 7$

The above can be rewritten as:

$\frac{log_2(x)}{4log_22} + \frac{log_2(x)}{2log_22} + log_2(x) = 7$

$log_22 = 1$

So, we have:

$\frac{log_2(x)}{4*1} + \frac{log_2(x)}{2*1} + log_2(x) = 7$

$\frac{1}{4}log_2(x) + \frac{1}{2}log_2(x) + log_2(x) = 7$

Multiply through by 4

$4(\frac{1}{4}log_2(x) + \frac{1}{2}log_2(x) + log_2(x)) = 7 * 4$

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

$7log_2(x) = 28$

Divide through by 7

$\frac{7log_2(x)}{7} = \frac{28}{7}$

$log_2(x) = 4$

Apply the following law of logarithm:

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

So, we have:

$x = 2^4$

$x = 16$