Problem:
For what value of x xx does
2 x + 2 x + 4 ( x 2 ) + 8 ( x 3 ) + 16 ( x 4 ) = 40 ? \log _{\sqrt{2}} \sqrt{x}+\log _{2} x+\log _{4}\left(x^{2}\right)+\log _{8}\left(x^{3}\right)+\log _{16}\left(x^{4}\right)=40 ?
log 2 ​ ​ x ​ + log 2 ​ x + log 4 ​ ( x 2 ) + log 8 ​ ( x 3 ) + log 1 6 ​ ( x 4 ) = 4 0 ?
Answer Choices:
A. 8 88
B. 16 161 6
C. 32 323 2
D. 256 2562 5 6
E. 1024 10241 0 2 4
Solution:
Rewriting each logarithm in base 2 gives
1 2 2 x 1 2 + 2 x + 2 2 x 2 + 3 2 x 3 + 4 2 x 4 = 40 \dfrac{\dfrac{1}{2} \log _{2} x}{\dfrac{1}{2}}+\log _{2} x+\dfrac{2 \log _{2} x}{2}+\dfrac{3 \log _{2} x}{3}+\dfrac{4 \log _{2} x}{4}=40
2 1 ​ 2 1 ​ log 2 ​ x ​ + log 2 ​ x + 2 2 log 2 ​ x ​ + 3 3 log 2 ​ x ​ + 4 4 log 2 ​ x ​ = 4 0
Therefore 5 2 x = 40 5 \log _{2} x=405 log 2 ​ x = 4 0 , so 2 x = 8 \log _{2} x=8log 2 ​ x = 8 , and x = n 256 x=n\boxed{256}x = n 2 5 6 ​ .
OR
For a ≠0 a \neq 0a î€ = 0 the expression 2 a ( x a ) = y \log _{2^{a}}\left(x^{a}\right)=ylog 2 a ​ ( x a ) = y if and only if 2 a y = x a 2^{a y}=x^{a}2 a y = x a . Thus 2 y = x 2^{y}=x2 y = x and y = 2 x y=\log _{2} xy = log 2 ​ x . Therefore the given equation is equivalent to 5 2 x = 40 5 \log _{2} x=405 log 2 ​ x = 4 0 , so 2 x = 8 \log _{2} x=8log 2 ​ x = 8 and x = 256 x= \boxed{256}x = 2 5 6 ​ .
The problems on this page are the property of the MAA's American Mathematics Competitions