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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
4 years, 8 months ago

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7 votes

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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.

Clifford Wilmot - 4 years, 8 months ago

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How you guys latexify the problem at brilliant??

Advitiya Brijesh - 4 years, 8 months ago

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brilliant.org/math-formatting-guide/

Clifford Wilmot - 4 years, 8 months ago

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@Clifford Wilmot no latex..

Advitiya Brijesh - 4 years, 8 months ago

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@Clifford Wilmot $\text{Thanks}$

Advitiya Brijesh - 4 years, 8 months ago

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@Advitiya Brijesh \text {Thanks}

Advitiya Brijesh - 4 years, 8 months ago

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@Advitiya Brijesh \(Thanks\)

Use \ (Thanks) without the first space.

Zi Song Yeoh - 4 years, 8 months ago

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@Zi Song Yeoh (Thanks)

Superman Son - 4 years, 7 months ago

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@Zi Song Yeoh It can also be \ (\text{Thanks}\ ) without spaces which would appear as \(\text{Thanks}\) :)

Nishanth Hegde - 4 years, 8 months ago

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@Nishanth Hegde $\ (\text{Thanks}\ )$

Advitiya Brijesh - 4 years, 8 months ago

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@Advitiya Brijesh $\(text\(THANKS)\)$

Superman Son - 4 years, 8 months ago

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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 \(x=0\)

Ian Mana - 4 years, 8 months ago

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Thank you for your reply but i cant take for granted that the solution is an integer.

Kon Tim - 4 years, 8 months ago

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Log of negative numbers is defined, as Calvin Sir has mentioned in some posts. Only log of 0 is not defined.

Sambit Senapati - 4 years, 8 months ago

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