We Complement Each Other.

Algebra Level 2

Let $f(x) = \dfrac{1}{1+2^{\lg x}} + \dfrac{1}{1+4^{\lg x}} + \dfrac{1}{1+8^{\lg x}}$ . Find the value of $f(x) + f \left(\dfrac{1}{x}\right)$ .

Notation: $\lg (\cdot) = \log_{10} (\cdot)$ denotes logarithm to the base 10.

The answer is 3.

2 solutions

Chew-Seong Cheong
Oct 30, 2019

$\begin{aligned} f(x) + \blue{f\left(\frac 1x\right)} & = \frac 1{1+2^{\lg x}} + \frac 1{1+4^{\lg x}} + \frac 1{1+8^{\lg x}} + \blue{\frac 1{1+2^{\lg \frac 1x}} + \frac 1{1+4^{\lg \frac 1x}} + \frac 1{1+8^{\lg \frac 1x}}} \\ & = \frac 1{1+2^{\lg x}} + \blue{\frac 1{1+2^{-\lg x}}} + \frac 1{1+4^{\lg x}} + \blue{\frac 1{1+4^{-\lg x}}} + \frac 1{1+8^{\lg x}} + \blue{\frac 1{1+8^{-\lg x}}} \\ & = \frac 1{1+2^{\lg x}} + \blue{\frac {2^{\lg x}}{2^{\lg x}+1}} + \frac 1{1+4^{\lg x}} + \blue{\frac {4^{\lg x}}{4^{\lg x}+1}} + \frac 1{1+8^{\lg x}} + \blue{\frac {8^{\lg x}}{8^{\lg x}+1}} \\ & = 1 + 1 + 1 = \boxed 3 \end{aligned}$

Yashas Ravi
Oct 30, 2019

We can substitute in $x=10$ (or any number, but $10$ is convenient) to obtain the answer of $3$ .

Isn't 1 a more convenient x

Kenny O. - 1 year, 7 months ago

Yes, any real number greater than 0 works :)

Yashas Ravi - 1 year, 7 months ago

Well, that would show that 3 is the only answer that could be true for all values of x. It would remain to be shown that 3 is the actual answer for all values of x.

Richard Desper - 1 year, 7 months ago

We Complement Each Other.

The answer is 3.

2 solutions

0 pending reports