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belka [17]
3 years ago
14

Log16^*+log4^*+log2^*=7​

Mathematics
1 answer:
ale4655 [162]3 years ago
6 0

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:

<em>If </em>log_ab = c<em> </em><em>Then </em>b = a^c<em></em>

So, we have:

x = 2^4

x = 16

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