Question

If $\log_{2} x+ \log_{4} x +\log_{16} x = \frac{21}{4}$, find x.

Answer 1

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Answer 2

Answer:

x = 8

Step-by-step explanation:

In the given question,

We have been provided an equation,

$log_{2}x+log_{4}x+log_{16}x=\frac{21}{4}$

Therefore, as the properties of the logarithms states that,

loga + logb + logc = log(abc)

Also,

$log_{a^{n}}b=\frac{1}{n}log_{a}b$

So, using the same properties we get,

$log_{2}x+log_{4}x+log_{16}x=\frac{21}{4}\\log_{2}x+log_{2^{2}}x+log_{2^{4}}x=\frac{21}{4}\\log_{2}x+\frac{1}{2}log_{2}x+\frac{1}{4}log_{2}x=\frac{21}{4}\\log_{2}x(1+\frac{1}{2}+\frac{1}{4})=\frac{21}{4}\\log_{2}x(\frac{7}{4})=\frac{21}{4}\\log_{2}x=3\\x=2^{3}\\x=8$

Therefore, on applying all the related formulas of the logarithms and solving the equation we finally get the simplified value of the equation as,

x = 8.

Therefore, the value of x = 8.