Question

Log2x+log4x+log16x=14

Answer 1

$\displaystyle \log_4x=\log_{2^2}x= \frac{1}{2}\log_2 x\\\\\log_{16}x=\log_{2^4}x= \frac{1}{4}\log_2 x$

Следовательно:

$\displaystyle \log_2 x+\frac{1}{2}\log_2 x+\frac{1}{4}\log_2 x=14\\\\ \frac{7}{4}\log_2x=14\\\\7\log_2x=56\\\\\log_2x=8\\\\2^8=x\\\\x=256$

Answer 2

Log₂ x + log₄ x + log₁₆ x = 14    x>0
log ₓⁿ t = 1/n * logₓ t
4log₂⁴ x + 2log₄² x + log₁₆ x = 14
4log₁₆ x + 2log₁₆ x + log₁₆ x = 14
7 log₁₆ x =14
log₁₆ x = 2
x = 16²= 256