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I spent hours trying to solve:

$$4^x + 1 = 2^{x+1}$$

  1. Can you guide me on how to solve this?

  2. How can I train myself to always find the right "trick" to solve such equations? Rather than just practicing... Is there a better way to always know how to approach such equations? Are there steps that will always lead me to the right way to approach? For example - "first always simplify" then "multiply" then... ?

Thanks!

Harish Chandra Rajpoot
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Marcus Wayne
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    Using same bases for the exponentiation as much as possible is often useful. Here use $4^x=2^{2x}=(2^x)^2$. – André Nicolas Aug 03 '15 at 00:08
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    Sadly, I don't know if there is a good technique for searching for quick solutions. "Search for a pattern" is the best I can manage. In this case, $4 = 2^2$ That has to help. And then...well, maybe a good substitution? But these techniques which work here, wouldn't work if, say, you changed that $2$ to a $3$. – lulu Aug 03 '15 at 00:09
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    I have a list of stuff that I go through that usually works: 1. Try to find a common base (see Andrés comment) 2. Try to find a substitution (see avid19's answer) 3. Black magic (see Harish's answer) – Vincent Aug 03 '15 at 10:38

4 Answers4

19

Let $u=2^{x}$. Then the equation becomes:

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

Solve this for $u$ then one gets $$x=\log_{2}(u)$$

Bhaskara-III
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8

We have, $$4^x+1=2^{x+1}$$ $$(2^2)^{x}+1=2\cdot2^{x}$$ $$(2^x){^2}-2\cdot 2^{x}+1=0$$ $$(2^x){^2}-2\cdot2^{x}+1=0$$ $$\underbrace{(2^x){^2}-2^{x}}\underbrace{-2^x+1}=0$$ $$2^x(2^x-1)-(2^x-1)=0$$ $$(2^x-1)(2^x-1)=0$$ $$(2^x-1)^2=0$$ $$2^x-1=0\iff 2^x=1$$ $$x\ln2=\ln1=0 $$$$\implies \bbox[5px, border: 2px solid #C0A000 ]{\color{red}{x=0}}$$

Harish Chandra Rajpoot
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5

$$4^x+1=2^{x+1}$$ $$2^{2x}+1=2^1\left(2^x\right)$$ $$\left(2^x\right)^2+1=2\left(2^x\right)$$ $$\left(2^x\right)^2-2\left(2^x\right)+1=0$$

Solve this as a quadratic equation in $2^x$:

$$2^x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(1)}}2$$

You can finish that, then use logarithms to find $x$ from $2^x$.

Rory Daulton
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    Once we got a canonical form af a quadratic function wrt. $2^x$ on the LHS, one may spot a well-known $(1, -2, 1)$ pattern of the coefficients and reduce LHS to $(2^x-1)^2$ without big–delta formula. – CiaPan Aug 03 '15 at 06:26
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Let $2^x=u$ $$4^x+1=2^{x+1}$$ $$u^2+1=2u$$ $$(u-1)^2=0$$ $$u=1$$ $$2^x=1$$ $$x\ln2=0$$ $$x=0$$