Solve for $x$ ; $ 2^x+4^x=8^x$

Question

I reduced the equation to

$1+2^x=2^{2x}$

It is easy to see that $x$ is irrational.I tried logarithms and failed to get an expression for $x$.Any help??

You're looking for two positive numbers whose sum is equal to their product, such that one is the square of the other. — Allawonder, Mar 24 '20 at 09:40

score 2 · Answer 1 · answered Mar 24 '20 at 08:18

2

Writing your expression as $(2^x)^2-2^x-1=0$ and solving we get $2^x=(1+\sqrt{5})/2$. Taking logarithms on both the sides you get the valueof x. Since $2^x$ cannot be -ve discard the other root of your equation.

answered Mar 24 '20 at 08:18

user159888

1,948
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score 1 · Answer 2 · answered Mar 24 '20 at 08:09

1

Suppose $y=2^x$, then from $1+2^x=2^{2x}$, we get $$1+y=y^2$$ could you finish?

answered Mar 24 '20 at 08:09

Qurultay

5,224

score 0 · Answer 3 · answered Mar 24 '20 at 08:09

0

Hint. Set $y=2^x$, solve for $y$. Then solve for $x$.

answered Mar 24 '20 at 08:09

SuperCiocia

576

score 0 · Answer 4 · answered Mar 24 '20 at 08:10

0

Hint: The equation is same as $y+y^{2}=y^{3}$ where $y=2^{x}$. Divide by $y$ and solve the quadratic equation in $y$.

answered Mar 24 '20 at 08:10

Kavi Rama Murthy

311,013

score 0 · Accepted Answer · answered Mar 24 '20 at 08:15

Let $z=2^x$, then $2^x+4^x=8^x \implies z+z^2=z^3 \implies z=0 or z^2-z-1$ So three roots of $z$ $0, \frac{1\pm \sqrt{5}}{2}$ as $z=2^x>0$ So only one real root is possible/ We have $$2^x=\frac{1+\sqrt{5}}{2}\implies x=\log_2 \left(\frac{1+\sqrt{5}}{2}\right). $$

Solve for $x$ ; $ 2^x+4^x=8^x$

5 Answers5