4

Let $\log x=\log 2+\log y$ and $2^x+8^y=4$ then find the $x$

My Try :

$$\log x=\log 2+\log y \\ \log x=\log 2y \\x=2y $$

So we have :

$$2^x+8^y=4\\2^{2y}+2^{3y}=4 \\t=2^y \\t^2+t^3=4$$

now what ?

Almot1960
  • 4,782
  • 16
  • 38

2 Answers2

7

from the first equation we have $$\log(x)=\log(2y)$$ and we get $$x=2x$$ plugging this in your second equation we have $$2^{2y}+8^y=4$$ this can be written as $$(2^y)^2+(2^y)^3=4$$ with $t=2^y$ we have to solve $$t^3+t^2-4=0$$ one solution is given by $$t=\frac{1}{3} \left(-1+\sqrt[3]{53-6 \sqrt{78}}+\sqrt[3]{53+6 \sqrt{78}}\right)$$ after the cubic formulas. This is the only real solution

Dr. Sonnhard Graubner
  • 95,283
  • 1
    Most of the substance of this answer is only repeating what the OP has clearly done already. Thus answer would've been a lot better if you had instead fleshed out the last part about the cubic formula. – Arthur Nov 27 '17 at 09:38
  • 1
    but with my answer the OP can see if his calculations are right, the cubic formula can be found in the internet – Dr. Sonnhard Graubner Nov 27 '17 at 09:40
1

I will assume you want a real solution. $t^2+t^3=4$ can be rewritten as $t^3+t^2-4=0$. Using the cubic formula, the real solution for $t$ is $$t=\frac{1}{3} \left(-1+\sqrt[3]{53-6\sqrt{78}}+\sqrt[3]{53+6 \sqrt{78}}\right)$$ We also know that $y=log_2(t)$ and that $x=2y$. So, using that, $$x=2\cdot log_2\left(\frac{1}{3} \left(-1+\sqrt[3]{53-6\sqrt{78}}+\sqrt[3]{53+6 \sqrt{78}}\right)\right)$$

Kyan Cheung
  • 3,184