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Just a simple equation!!!

Hello, i would like to provide me a full solution for this equation 2^x + 3^x = 2 I know that the answer is x = 0 but i would like to tell me how i solve it step by step I wonder also if this is a right way to solve it 2^X + 3^X = 1 + 1 => 2^X + 3^X = 2^0 + 3^0 => 2^X = 2^0 AND 3^x = 3^0 so x = 0

Note by Kon Tim
3 years, 11 months ago

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$$\frac{d}{dx}(2^x+3^x)=ln(2)2^x+ln(3)3^x$$.

$$ln(2)2^x+ln(3)3^x > 0$$ so the gradient of $$2^x+3^x$$ is always positive, so $$2^x+3^x$$ must be increasing. Hence $$2^x+3^x$$ can only cross the line $$y=2$$ once, so $$2^x+3^x=2$$ can only have one solution. By observation, $$x=0$$ is a solution, and so must be the only solution. · 3 years, 11 months ago

How you guys latexify the problem at brilliant?? · 3 years, 11 months ago

brilliant.org/math-formatting-guide/ · 3 years, 11 months ago

no latex.. · 3 years, 11 months ago

$\text{Thanks}$ · 3 years, 11 months ago

\text {Thanks} · 3 years, 11 months ago

$$Thanks$$

Use \ (Thanks) without the first space. · 3 years, 11 months ago

(Thanks) · 3 years, 11 months ago

It can also be \ (\text{Thanks}\ ) without spaces which would appear as $$\text{Thanks}$$ :) · 3 years, 11 months ago

$\ (\text{Thanks}\ )$ · 3 years, 11 months ago

$$$text\(THANKS)$$$ · 3 years, 11 months ago

this is just my own step but you can ask if there is something you want to add :)

$$2^{x} + 3^{x} = 2$$

Transposing $$2^{x}$$ to the right side

$$3^{x} = 2-2^{x}$$

by the use of logarithm

$$log_{3}(2-2^{x}) = x$$

by using the law of logarithm, $$(2 - 2^{x})$$ must be greater than zero because you cannot log the number 0 or any negative number, meaning $$x<1$$

so the only integer that satisfies the condition is 0 so $$x=0$$ · 3 years, 11 months ago

Thank you for your reply but i cant take for granted that the solution is an integer. · 3 years, 11 months ago