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I'm drawing a total blank as to how to solve this for x ...

$$ 1.1^x + 1.5^x = 3.46 $$

I thought I could differentiate and substitute the first derivative back into the equation, but I must be rustier than I thought because I'm it's not working out correctly.

Incidentally, the answer is $x=2$.

Any thoughts on how to solve this type of equations would be very much appreciated.

Kindest regards,

Jack

String
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Jack CL
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2 Answers2

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Well, if you already see that $x=2$ is a solution, then you only have to add to this observation that the function $$x\ \mapsto\ (1.1)^x+(1.5)^x$$ is strictly increasing, so, in particular it is one-to-one, i.e. takes a value at most once, which proves that $x=2$ is the only solution.

Berci
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I'd use Newton's Method to compute this in 2-3 steps after setting up the correct formula:

$$ x_{n+1} = x_n - \frac {1.1^x + 1.5^x-3.46}{0.0953102 \cdot 1.1^x + 0.405465 \cdot 1.5^x} $$

where $x_0 = 1$.

I don't believe there's an analytical way to solve for $x$ efficiently, unless you go crazy with a Taylor series representation of $f(x) = 1.1^x + 1.5^x$ centered at $x=0.$

Of course, you can always plug 'n' chug for small values..

daOnlyBG
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  • How about expressing $1.5$ as $1.1^{log_{1.1}{1.5}}$ and then using Lambert W? It would not be so crazy and the result would be a closed formula. – peterh Jun 02 '19 at 01:06