Intuition will work here?

Algebra Level 4

$\large 2^x+(0.5)^{2x-3}-6(0.5)^x=1$

How many real values of $x$ satisfy the equation above?

Bonus question: Determine the value(s) of $x$ .

1 More than 2 2 0

1 solution

Jaydee Lucero
Jul 30, 2015

$\begin{aligned} 2^x + (0.5)^{2x-3} - 6(0.5)^x &= 1 \\ 2^x + \frac{1}{2^{2x-3}} - 6 \cdot \frac{1}{2^x} &= 1 \\ 2^x + \frac{8}{2^{2x}} - 6 \cdot \frac{1}{2^x} &= 1 \\ 2^{3x} + 8 -6 \cdot 2^x &= 2^{2x} \\ 2^{3x}-2^{2x}-6\cdot 2^x + 8 &= 0\end{aligned}$ Letting $y = 2^x$ , the last equation becomes $y^3-y^2-6y+8=0$ By synthetic division, we find that $y = 2$ is a root. The reduced equation thus becomes $y^2+y-4=0$ , and by quadratic formula, the two other values are $y=\frac{-1\pm\sqrt{17}}{2}$ .

However, take note that we have the substitution $y=2^x$ , thus $y \ge 0$ for all $x$ . Hence, the value $y=\frac{-1-\sqrt{17}}{2}$ is not acceptable.

Thus, there are only $\fbox{2}$ values of $x$ that satisfy the equation. These are $y=2^x=2 \Longrightarrow x=1$ and $y=2^x=\frac{-1+\sqrt{17}}{2} \Longrightarrow x = \log_{2}{\frac{-1+\sqrt{17}}{2}}$

I did exactly the same :D

Romeo Gomez - 5 years, 10 months ago

This is a very silly question but would you explain why the exponents weren't used for 1? as in 0.5 has been broken down to 1/2 bu only on 2 the exponent is used? I hope my question is understandable :P

이채 린 - 5 years, 10 months ago

For all real $x$ , $1^x = 1$ . So, for example,

$(0.5)^x=\left( \frac{1}{2} \right)^x = \frac{1^x}{2^x} = \frac{1}{2^x}$

Jaydee Lucero - 5 years, 10 months ago

now it's clear to me thank you :D

이채 린 - 5 years, 10 months ago

Intuition will work here?

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