An Interesting Exponential Equation | Math Olympiad Prep!

[@gregevgeni1864]

Let (1) the initial equation.
But (√44 + √144)^x = (2√11 + 12)^x = (1+√11)^(2x) = [(1+√11)^x]^2.
[(1+√11)^x]^2 and the (1)
[(1+√11)^x]^2 = x^2
[(1+√11)^x + x]•[(1+√11)^x - x] = 0 .
.....

[@mohammedsaysrashid3587]

It was a wonderful introduction and clearly explaining thank you Sir for sharing....x=-0,489

[@SidneiMV]

√44 + √144 = 12 + 2√11 = (1 + √11)²
*1 + √11 = a* => (1 + √11)² = a²
a^2x = x²
x = ± a^x
x = a^x
lnx = xlna
(1/x)lnx = lna
(1/x)ln(1/x) = ln(1/a)
ln(1/x) = W[ln(1/a)]
1/x = e^W[ln(1/a)]
x = e^-W[ln(1/a)]
*x = W[ln(1/a)]/ln(1/a)*
x = - a^x => a^x = -x
u = -x => x = -u
u = a^(-u)
-ulna = lnu
(1/u)lnu = -lna
(1/u)ln(1/u) = lna
ln(1/u) = W(lna)
1/u = e^W(lna)
u = e^-W(lna) = [W(lna)]/lna
*x = - [W(lna)]/lna*

[@RashmiRay-c1y]

Note that sqrt(144) + sqrt(44) = 12+2sqrt(11) = (sqrt(11) + 1)^2. So, (sqrt(11) + 1)^(2x) = x^2 > 2ln x = 2xln(sqrt(11)+1)> 1/x ln(1/x) = - ln(sqrt(11)+1) > ln(1/x) e^(ln(1/x)) = - ln(sqrt(11)+1) > ln(1/x) = W(-ln(sqrt(11)+1)) > x = e^[- W(-ln(sqrt(11)+1))] = 0.035 - 1.05i. So, x = 0.035 - 1.05i. If we solve x^2 = (sqrt(11)+1)^2x graphically, we get x = -489.

[@xualain3129]

You are right. There is no real solution for W(-ln(sqrt(11)+1)).

[@RashmiRay-c1y]

@@xualain3129 I tried to evaluate the appropriate value for the Lambert W function on Wolfram Alpha and got a complex number as the answer. However, instead of taking that route, if we solve (sqrt(11)+1)^(2x) = 18.7^x = x^2 graphically, we do get a real solution, x=-0.489. There must be another branch of the W function that Wolfram Alpha is missing. I wonder why.

[@xualain3129]

@@RashmiRay-c1yThank you for your quick response. After close review, I guess I found out the other solution we neglected.
With (sqrt11)+1)^(2*x)=x^2
We should get two solutions.
(sqrt(11)+1)^x=x which leads to no real solution.
(sqrt(12)+1)^x=-x which can be eventually deduced to
W(ln(sqrt(11)+1)=-x*ln(sqrt(11)+1)
W(1.4624)=-x*1.4624
0.71523=-x*1.4624
x=-0.489
That is the real solution we looked for.

[@RashmiRay-c1y]

@@xualain3129 Excellent!

[@ericerpelding2348]

Can this be solved without using the W function?
I asked this because some algebra classes do not cover ro mention that function.

[@RashmiRay-c1y]

Yes. If we solve x^2 = (sqrt(11)+1)^2x graphically, we get x = -489.

[@infyGyan]

We may use the Newton raphson method too to get the solutions.

[@moeberry8226]

@@RashmiRay-c1yx cannot equal -489 that’s impossible your math is off.

[@dorkmania]

How did you get the value of e^-W[-ln(√11 + 1)] as -0.4891?