A Cool Exponential Equation

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[@Qermaq]

I noticed that 2^a + 2^b = 24, and one way to do that is a = 3 and b = 4, or the other way around. Since the second exponent (b) has a larger numerator, I figured it was probably the larger value, so I set the first exponent equal to 3. The "got lucky" method!

[@tejpalsingh366]

2^4+ 2^3. Now get easily n fast

[@duhdinduh]

x=2

[@phill3986]

😊😊😊👍👍👍

[@SidneiMV]

(2x - 1)/(x - 1) = 2 + 1/(x - 1)
(3x - 2)/(x - 1) = 3 + 1(x - 1)
2^[1/(x - 1)] = u
4u + 8u = 24 => u = 2
2^[1/(x - 1)] = 2
1/(x - 1) = 1 => *x = 2*

[@nasrullahhusnan2289]

[2^(2x-1)]+2^(3x-2)=24
Divide by 2³:
[2^(2x-4)]+2^(3x-5)=3
[2^(2x-4)]-1+[2^(3x-5)]-2=0
[2^(x-2)]²-1=2-[2^(3x-5)]
=2[1-{2^(3x-6)}]
=-2[2^(x-2)]²-1]
u=-2u --> u=0
[2^(x-2)]²-1=0
--> 2^(x-2)=±1
2^(x-2)=1 as 2^(x-2)>0
x-2=0 --> x=2

[@josepherhardt164]

Before even viewing: Is this another hidden quadratic? (Scribbles furiously on paper.) No. Darn, I'm SO disappointed! :(

[@mystychief]

Simple method. Equation 1: 2^a+2^b=(2^3)3. Divide by 2^3 gives 2^(a-3)+2^(b-3)=3. What if it's simple and this second equation is 1+2=3 (b probably being greater than a). That means the exponents of 2 are then probably respectively 0 and 1 or a-3=0 and b-3=1. After filling in the a and b again and solving for x both give the solution x=2. Solved.