Find all the solutions of the equation

[mathmari]

Hey!

I have to find all the solutions of the equation
$$3^x+4^x=5^x$$

I have done the following:
$$3^x+4^x=5^x \Rightarrow (\frac{3}{5})^x+(\frac{4}{5})^x=1$$
$$f(x)=(\frac{3}{5})^x+(\frac{4}{5})^2-1 \Rightarrow f'(x)=(\frac{3}{5})^x \ln{(\frac{3}{5})}+(\frac{4}{5})^2 \ln{(\frac{4}{5})}$$

Is it true that $f'(x)<0, \forall x \in \mathbb{R}$?? (Wondering)

If so, we would know that $f(x)$ has at most one solution.

How could I continue?? (Wondering)

 

[chisigma]

Hey!

I have to find all the solutions of the equation
$$3^x+4^x=5^x$$

I have done the following:
$$3^x+4^x=5^x \Rightarrow (\frac{3}{5})^x+(\frac{4}{5})^2=1$$
$$f(x)=(\frac{3}{5})^x+(\frac{4}{5})^2-1 \Rightarrow f'(x)=(\frac{3}{5})^x \ln{(\frac{3}{5})}+(\frac{4}{5})^2 \ln{(\frac{4}{5})}$$

Is it true that $f'(x)<0, \forall x \in \mathbb{R}$?? (Wondering)

If so, we would know that $f(x)$ has at most one solution.

How could I continue?? (Wondering)

Your idea is good but the correct procedure is to write...

$\displaystyle f(x) = (\frac{3}{5})^{x} + (\frac{4}{5})^{x} = 1\ (1)$

Because $\displaystyle f^{\ '} (x) < 0$ for all x , the equation has only one solution and it is easy o see that this solution is x=2...Kind regards

$\chi$ $\sigma$

 

[mathmari]

Your idea is good but the correct procedure is to write...

$\displaystyle f(x) = (\frac{3}{5})^{x} + (\frac{4}{5})^{x} = 1\ (1)$

Because $\displaystyle f^{\ '} (x) < 0$ for all x , the equation has only one solution and it is easy o see that this solution is x=2...Kind regards

$\chi$ $\sigma$

Do we know that $\displaystyle f^{\ '} (x) < 0$ for all x, because of the fact the $\ln{y}$ is negative in this case, and $y^x$>0, $y=\frac{3}{5},\frac{4}{5}$? (Wondering)

 

[Prove It]

Do we know that $\displaystyle f^{\ '} (x) < 0$ for all x, because of the fact the $\ln{y}$ is negative in this case, and $y^x$>0, $y=\frac{3}{5},\frac{4}{5}$? (Wondering)

$\displaystyle \begin{align*} f(x) &= \left( \frac{3}{5} \right) ^x + \left( \frac{4}{5} \right) ^x - 1 \\ \\ f'(x) &= \left( \frac{3}{5} \right) ^x \ln{ \left( \frac{3}{5} \right) } + \left( \frac{4}{5} \right) ^x \ln{ \left( \frac{4}{5} \right) } \end{align*}$

Since $\displaystyle \begin{align*} \left( \frac{3}{5} \right) ^x \end{align*}$ and $\displaystyle \begin{align*} \left( \frac{4}{5} \right) ^x \end{align*}$ are both always positive, and $\displaystyle \begin{align*} \ln{ \left( \frac{3}{5} \right) } \end{align*}$ and $\displaystyle \begin{align*} \ln{ \left( \frac{4}{5} \right) } \end{align*}$ are both negative, we can conclude that the derivative is always negative.

 

[chisigma]

Do we know that $\displaystyle f^{\ '} (x) < 0$ for all x, because of the fact the $\ln{y}$ is negative in this case, and $y^x$>0, $y=\frac{3}{5},\frac{4}{5}$? (Wondering)

Is...

$\displaystyle f(x) = e^{x\ \ln \frac{3}{5}} + e^{x\ \ln \frac{4}{5}} \implies f^{\ '} (x) = \ln \frac{3}{5}\ e^{x\ \ln \frac{3}{5}} + \ln \frac{4}{5}\ e^{x\ \ln \frac{4}{5}}\ (1)$

... and because $\displaystyle \ln \frac{3}{5}$ and $\displaystyle \ln \frac{4}{5}$ are both negative, the $\displaystyle f^{\ '} (x)$ is also negative...

Kind regards$\chi$ $\sigma$

 

[mathmari]

$\displaystyle \begin{align*} f(x) &= \left( \frac{3}{5} \right) ^x + \left( \frac{4}{5} \right) ^x - 1 \\ \\ f'(x) &= \left( \frac{3}{5} \right) ^x \ln{ \left( \frac{3}{5} \right) } + \left( \frac{4}{5} \right) ^x \ln{ \left( \frac{4}{5} \right) } \end{align*}$

Since $\displaystyle \begin{align*} \left( \frac{3}{5} \right) ^x \end{align*}$ and $\displaystyle \begin{align*} \left( \frac{4}{5} \right) ^x \end{align*}$ are both always positive, and $\displaystyle \begin{align*} \ln{ \left( \frac{3}{5} \right) } \end{align*}$ and $\displaystyle \begin{align*} \ln{ \left( \frac{4}{5} \right) } \end{align*}$ are both negative, we can conclude that the derivative is always negative.

Is...

$\displaystyle f(x) = e^{x\ \ln \frac{3}{5}} + e^{x\ \ln \frac{4}{5}} \implies f^{\ '} (x) = \ln \frac{3}{5}\ e^{x\ \ln \frac{3}{5}} + \ln \frac{4}{5}\ e^{x\ \ln \frac{4}{5}}\ (1)$

... and because $\displaystyle \ln \frac{3}{5}$ and $\displaystyle \ln \frac{4}{5}$ are both negative, the $\displaystyle f^{\ '} (x)$ is also negative...

Kind regards$\chi$ $\sigma$

Great! Thank you both! (Happy)

To find that $x=2$ is the solution, is there a method? Or do we have to check diverses $x$ and then we see that $x=2$ satisfies the relation?

 

[Prove It]

Great! Thank you both! (Happy)

To find that $x=2$ is the solution, is there a method? Or do we have to check diverses $x$ and then we see that $x=2$ satisfies the relation?

I can't think of a method to solve this type of equation analytically. But you should know some basic Pythagorean Triples...

 

[mathmari]

I can't think of a method to solve this type of equation analytically. But you should know some basic Pythagorean Triples...

Ok...Thanks! (Smile)