Solving A Cool Exponential Equation

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[@moeberry8226]

X=2 and since this is always continuous and after dividing by 3^x we have an increasing function equal to a decreasing function so there is only 1 solution.

[@Skyler827]

I don't like this. You apply some helpful transformations, but the solution is just guess and check. There is no reason to expect a real world problem to have such a nice, guessable solution. If the answer were something more complicated, I don't know how I would solve it.

[@redpalkia8220]

that's my exact problem with this type of problem, a^x + b^x = c has seemingly no possible way of solving other than through graphing

[@farhansadik5423]

@@redpalkia8220 you can use numerical approximations, but that's it

[@quantumbuddha777]

I would use Newton-Raphson if it were a real world problem I needed to solve

[@aplusbi]

This is not real world! 😜

[@simonsigey2578]

No need to divide through by any number

[@honestadministrator]

f ( x) = 3 ^ x - ( 2 √2) ^x - 1 has a zero. at x = 2
d/dx ( f ( x))
= 3 ^ x * ln (3) - ln ( 2 √2) * ( 2 √2) ^x
HERE BY f ( x) is a monotonically imcreasing function

[@phill3986]

😊😊😊👍👍👍

[@simonsigey2578]

Solve like difference of 2 squares

[@SyberMath]

How?

[@2012tulio]

It will take you nowhere

[@georgegeorgopoulos1861]

Not an interesting problem. There is no way to solve this problem analytically. You just use a guess and check algorithm which is basically what a computer numerical algorithm would do.

[@DonEnsley-mathdrum]

problem
3ˣ - (2√2)ˣ = 1
let
u = 3ˣ
and
v = (2√2)ˣ
We know u-v = 1. That is the main equation with which we are dealing.
Therefore u-v-1 = 0
Set x = x by setting ln u/ ln 3 = ln v / ln (2√2)
ln u/ ln 3 = ln v / ln (2√2)
Replace u = v+1
ln (2√2)/ ln 3 = ln v / ln (v+1)
ln v / ln (v+1) = ln (2√2)/ ln 3
The only solution is v=8 so
ln v / ln (v+1) =
ln 8 / ln 9 =
3 ln 2 / 2 ln 3 =
ln (2³ᐟ²)/ln 3
u= 9=3ˣ
v = 8= (2√2)ˣ
answer
x = 2

[@fadetoblah2883]

That approach does not capture the one obvious solution x=2; this should be a clue that the method is flawed. Note that when reaching (u+v)(u-v-1)=0, it cannot be assumed that both u+v=0 and u-v-1=0 are true at the same time, which is exactly what is being done here when using v+1 as a substitute for u in both factors, because u=v+1, the proposed substitution, is just an algebraic equivalent of u-v-1=0, the second factor. This equivalency, by the way is what causes u-v-1=0 to turn into v+1-v-1=0, and then 0=0 (no new information gained here) after replacing u with v+1.

[@DonEnsley-mathdrum]

@@fadetoblah2883 thanks. Have modified comment using a different approach. It now reflects x=2 as the only solution.

[@rakenzarnsworld2]

x = 2

[@mega_mango]

hi

[@scottleung9587]

I also got x=2 as the only solution.