How do you solve this tricky exponential?

Question

This problem is tricky because of differing bases and a squared variable.

3^x^2 = 6^(x+4)

Can this equation be solved for x without using a calculator, if it is shown in terms of logarithms?

Answer 1

did you you mean 3^(x²) = 6^(x + 4)?

take log for both sides
log 3^(x²) = log 6^(x + 4)

exponent rule
x^2 log3 = (x + 4) log6

distribute
x^2 log3 = xlog6 + log6^4

x^2 log3 = xlog6 + log1296

x^2 log3 - xlog6 - log1296 = 0

a = log3
b = -log6
c = -log1296

use quadratic formula and you'll get x = -1.865711 and 3.4966404

Answer 2

Take the natural log of the both sides. You get X^2 *ln3=(x+4)*ln6. Moving the unknown to one side, you'll have X=x^2/(x+4)=log(3)6 (log 6 of base 3)=1.63093. You can make it a quadratic equation x^2-[log(3)6]*x-4*[log(3)6]=0. X=[3.4966, -1.8657]

Answer 3

3^x^2 = 6^(x+4)
ln 3x^2= ln 6^(x+4)
x^2ln3 = (x+4)ln6
x^2/(x+4) = ln6/ln3
x^2 = x(ln6/ln3) + 4(ln6/ln3)
x^2 - ln6/ln3)x - 4* ln(6)/ln3
x=[ln6/ln3+/-sqrt((ln6/ln3)^2-4(1)(-4(ln6/ln3))]/2
The above is an exact solution.

Without a calculator, it would be difficult to obtain an approximate answer because you would not know the value of ln6/ln3.