Question

A friend gave me a riddle... which involves solving for x in the equation: 2^x=3^(x/2)+1.
Apparently this can be solved using simple algebra, but I cannot seem to figure this out. Can anyone help?

Answer 1

Is this your question?
\[\huge {2^{x}=3^{\frac{x}{2}}+1}\]

Answer 2

Yes. I know the answer is two, but I am trying to solve this algebraically.

Answer 3

I was thinking about using the Lambert W function, but I am not familiar with that sort of thing.

Answer 4

It can be written as
\[\huge{e^{\ln {2^x}}=e^{\ln {3^{\frac{x}{2}}}}+1}\]
Let's try by using expandion of e^x

Answer 5

no it can't ash.
If u raise both sides to e, u'll get:
e^(ln2^x) = e*e^(ln(3^(x/2)))

Answer 6

Bahrom I've not raised to e
I've used the property
\[a^{log_{a} b}=b\]

Answer 7

but.. but u can't do that lol. how?

Answer 8

That's a property of logarithms
\[ \log_{2} 4= 2\]
so
\[2^{\log_{2} 4}=2\]

Answer 9

Sorry 4

Answer 10

wait
2^x = 3^(x/2) + 1
taking ln on both sides gives:
ln(2^x) = ln(3^(x/2) + 1) u can't distribute ln in on the right.

Answer 11

Nope

Answer 12

yea so how did u get that 1 by itself on the right, that's impossible.

Answer 13

Simple algebra? It's clear by simple algebra that x = 2 is a solution.

What's not so obvious is that it's the only solution. To show that, you need some calculus:

Consider the function
f(x) = 2^x - ( 3^(x/2) + 1 )

Then f is monotonically increasing (i.e., f' > 0) and hence x = 2 can be the only solution.

Answer 14

you can put it in the form
(3+1)^y = 3^y+1

where y= x/2

expand the binomial, cancel the highest and lowest term
only solution to give zero is (0 choose 1) so y=1, and x=2

Answer 15

Thanks for the help.

Answer 16

@phi: that assumes that y is an integer, which it need not be.

Answer 17

good point. Insert, for all integer y...
If you want the fractional solutions, good luck finding them.