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

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

\(ln(2)2^x+ln(3)3^x > 0\) 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, 1 month ago

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– Advitiya Brijesh · 4 years, 1 month ago

How you guys latexify the problem at brilliant??Log in to reply

– Clifford Wilmot · 4 years, 1 month ago

brilliant.org/math-formatting-guide/Log in to reply

– Advitiya Brijesh · 4 years, 1 month ago

no latex..Log in to reply

– Advitiya Brijesh · 4 years, 1 month ago

$\text{Thanks}$Log in to reply

– Advitiya Brijesh · 4 years, 1 month ago

\text {Thanks}Log in to reply

Use \ (Thanks) without the first space. – Zi Song Yeoh · 4 years, 1 month ago

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– Superman Son · 4 years, 1 month ago

(Thanks)Log in to reply

– Nishanth Hegde · 4 years, 1 month ago

It can also be \ (\text{Thanks}\ ) without spaces which would appear as \(\text{Thanks}\) :)Log in to reply

– Advitiya Brijesh · 4 years, 1 month ago

$\ (\text{Thanks}\ )$Log in to reply

– Superman Son · 4 years, 1 month ago

$\(text\(THANKS)\)$Log in to reply

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, 1 month ago

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– Kon Tim · 4 years, 1 month ago

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

– Sambit Senapati · 4 years, 1 month ago

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