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I'm trying to solve the problem, but I didn't get the way to do it:

$$3^x + 4^x = 15$$

I tried the $\ln$-way, but it didn't help. I'm wondering how to find $x$ in this case

Blue
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Shaka Zulo
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  • It doesn´t look like that the equation can be solved algebraically. It would be easier if the equation was $3^x+4^x=25$. – callculus42 Oct 06 '19 at 14:43
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    Algebra of elementary functions won't help you here. There is exactly one real solution. If you manage to find it somehow good, otherwise less good. –  Oct 06 '19 at 14:43
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    The logarithm doesn't work because log(a+ b) does not simplify. The best way I could think to do this is to note that 3^1+ 4^1= 7< 15 and 3^2+ 4^3 25> 15. So x is between 1 and 2,. Trying x= 1.5, 3^1.5+ 4^1.5= 13.20< 15 so x is between 1.5 and 2. Halfway between is 1.75. 3^1.75+ 4^1.75= 18.15> 15 so x is between 1.5 and 1.75. Continue until you have sufficient accuracy. – user247327 Oct 06 '19 at 14:50
  • The first few powers of $3$ are $3,9,27,\dots$ and the first few powers of $4$ are $4,16,64,\dots.$ It's easy to see that there is no integer solution. Also, this reveals that the solution is between $1$ and $2,$ closer to the former. – Allawonder Oct 06 '19 at 14:57

2 Answers2

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The function on the left is an increasing function of $x$ and will have just one solution somewhere between $x=1$ and $x=2$. There is no way to "solve" the equation with a formula. You will need numerical methods.

Wolfram Alpha says the answer is about $x = 1.60057$.

Ethan Bolker
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As said in comments and answers, the is no analytical solution of the problem which means that you need some numerical method for solving the equation.

If you plot $$f(x)=3^x+4^x-15$$ it is very stiff while $$g(x)=\log(3^x+4^x)-\log(15)$$ looks very linear. This would help.

On the other side, you could ntice that $f(x)$ is bracketed by $2 \times 3^x-15$ and $2 \times 4^x-15$ which means that the solution is such that $$\frac{\log \left(\frac{15}{2}\right)}{\log (4)}<x <\frac{\log \left(\frac{15}{2}\right)}{\log (3)}$$ which are $1.45$ and $1.83$.

So, using Newton method, let us start at the midpoint of the interval; the successive iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.6437445 \\ 1 & 1.6005827 \\ 2 & 1.6005684 \end{array} \right)$$

Claude Leibovici
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