How do you solve #4^x + 6(4^-x) = 5#?

1 Answer
Jim H
Aug 5, 2015

#x=1/2" "# or #" "x=ln3/ln4#

Explanation:

Don't be too concerned about what you should do to solve the problem.
Focus on what you could do and on the question, "Would that help?"

#4^x + 6(4^-x) = 5#

Well, perhaps I could write 6 as 4 to some power. But i can add 4 to a powe plus 4 to another power, so that doesn't seem helpful.

Think of something else we could do.
Negative exponents can be a nit trick, so let's rewrite:

#4^x + 6/4^x = 5#

I've had good luck in other problems with clearing fraction, so let's do that.

#(4^x)^2 + 6 = 5 (4^x)#

Does that help? It's not really clear yet, but I do see a square, and a constant and a constant times the thing that is squared. That sound like a quadratic to me.

#(4^x)^2 -5(4^x)+ 6 = 0" "# Yes. We have:

#u^2 - 5u+6=0" "# with #4^x# rather than #u#.

#(u-2)(u-3)= 0#

#u = 2" "# or #" "u=3#

#4^x = 2" "# or #" "4^x=3#

#x=1/2" "# or hmmmm. The best I can do is something like

#x=1/2" "# or #" "x=log_4 3#

If I'd rather have #ln# than #log_4#, I'll write:

#x=1/2" "# or #" "x=ln3/ln4#

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