An Exponential Problem

[AvgStudent]

I've come across this little equation that appears easy but giving me a hard time. Note that this problem isn't homework nor has an impact on my grade in any way. Does anyone have an idea how to approach it?
[math]5^x-3^x=16[/math]
Solve for all [imath]x \in \R[/imath].

EDIT:
I've graphed the equation, and it appears that 2 is the only solution. If this help.

 

[topsquark]

I've come across this little equation that appears easy but giving me a hard time. Note that this problem isn't homework nor has an impact on my grade in any way. Does anyone have an idea how to approach it?
[math]5^x-3^x=16[/math]
Solve for all [imath]x \in \R[/imath].

EDIT:
I've graphed the equation, and it appears that 2 is the only solution. If this help.

Graphing sounds good. The only other way I can think of is to note:
[imath]5^x - 3^x = 16[/imath]

[imath]5^x = 3^x + 4^2[/imath]
and note that this could be seen as a Pythagorean triple.

-Dan

 

[Dr.Peterson]

I've come across this little equation that appears easy but giving me a hard time. Note that this problem isn't homework nor has an impact on my grade in any way. Does anyone have an idea how to approach it?
[math]5^x-3^x=16[/math]
Solve for all [imath]x \in \R[/imath].

EDIT:
I've graphed the equation, and it appears that 2 is the only solution. If this help.

Without actually graphing, I can imagine the LHS as the dstance between the graphs of [imath]3^x[/imath] and [imath]5^x[/imath], which will clearly be increasing (for positive x, which is necessary for the LHS to be positive), so there will be only one solution. Then just trying a few values of x (1, 2, ...) we quickly find the answer.

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If the answer were not an integer, we could try something like a bisection method. I can't see a more algebraic method, unless someone knows a very special trick.