Difficult exponential equation $6^x-3^x-2^x-1=0$

$6^x-3^x-2^x-1=0$ equivalent $1=\frac 1 {2^x}+\frac 1 {3^x}+\frac 1 {6^x}$

The left side is constant while the right side is decreasing. There can only be a solution at most, and $x=1$ is perfect for the job.


I quite liked your question so thought this might be of assistance next time you face a question similar to this one you can employ some mathematical software to help for e.g. Maple. You can rewrite the equation to look like this $$ f(x) = g(x) $$ where $f(x)= 6^x - 3^x -2^x$ and $g(x)=1$ is a constant function. So it's an intersection problem. The graph is below.

enter image description here

Clearly, they intersect at $x=1$ or you can use the following code to be $100\%$ sure

fsolve({f(x) = g(x)}, {x}, x = -1 .. 2)

where $f(x)$ and $g(x)$ are defined as above.

Tags:

Exponential Function

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