log16^*+log4^*+log2^*=7​

Answer:

[tex]x = 16[/tex]

Step-by-step explanation:

Given

[tex]log_{16}(x) + log_4(x) + log_2(x) = 7[/tex]

Required

Solve for x

[tex]log_{16}(x) + log_4(x) + log_2(x) = 7[/tex]

Change base of 16 and base of 4 to base 2

[tex]\frac{log_2(x)}{log_2(16)} + \frac{log_2(x)}{log_2(4)} + log_2(x) = 7[/tex]

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

[tex]\frac{log_2(x)}{log_2(2^4)} + \frac{log_2(x)}{log_2(2^2)} + log_2(x) = 7[/tex]

The above can be rewritten as:

[tex]\frac{log_2(x)}{4log_22} + \frac{log_2(x)}{2log_22} + log_2(x) = 7[/tex]

[tex]log_22 = 1[/tex]

So, we have:

[tex]\frac{log_2(x)}{4*1} + \frac{log_2(x)}{2*1} + log_2(x) = 7[/tex]

[tex]\frac{1}{4}log_2(x) + \frac{1}{2}log_2(x) + log_2(x) = 7[/tex]

Multiply through by 4

[tex]4(\frac{1}{4}log_2(x) + \frac{1}{2}log_2(x) + log_2(x)) = 7 * 4[/tex]

[tex]log_2(x) + 2}log_2(x) + 4log_2(x) = 28[/tex]

[tex]7log_2(x) = 28[/tex]

Divide through by 7

[tex]\frac{7log_2(x)}{7} = \frac{28}{7}[/tex]

[tex]log_2(x) = 4[/tex]

Apply the following law of logarithm:

If [tex]log_ab = c[/tex] Then [tex]b = a^c[/tex]

So, we have:

[tex]x = 2^4[/tex]

[tex]x = 16[/tex]

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