2^x times 3^x = 8 solve for x?

2^x * 3^x = 8
3^x = 8/(2^x)
3^x = (2^3)/(2^x)
3^x = 2^(3 - x)
log(3^x) = log[2^(3 - x)]
xlog(3) = (3 - x)log(2)
xlog(3) = 3log(2) - xlog(2)
xlog(3) + xlog(2) = log(2^3)
x[log(3) + log(2)] = log(8)
x = log(8)/[log(3) + log(2)]
x = log(8)/log(3 * 2)
x = log(8)/log(6)
x ≈ 1.1606

2^x * 3^x = 8
6^x = 8
take logs of both sides
log(6^x) = log8
x log6 = log 8
x = log8 / log 6
x = 1.16 (2dp)

I wonder who is giving out the thumbs down? This answer is correct and can easily be confirmed:

2^1.16 * 3^1.16 = 8

(2^x) (3^x) = 8
6^x = 8
x = log6(8) ....... log6 means log base 6
= log6((2)³)
= 3 log6(2)

Answer: x = 3 log6(2) ...... approx. 1.160558422 decimal.

Just to confirm:
2^(3 log6(2)) 3^(3 log6(2))
= 6^(3log6(2)) .......... (x^a)(y^a) = (xy)^a
= (6^(log6(2)))³ ........ x^(yz) = (x^y)^z
= (2)³ ........................ b^(logb(x)) = x
= 8

2^x * 3^x = 8
6^x = 8
x = log (base 6) of 8
x = log 8 ÷ log 6

or

x = ln 8 ÷ ln 6

Convert to logs:
x log 2 + x log 3 = log 8
x (log 2 + log 3) = log 8
Divide both sides by the quantity in parentheses
x = log 8 /(log 2 + log 3)
x = 1.16

Yes, it's logs ! xlog2 + xlog3 = log8 so x = log8/(log2 + log3) = 1.16 by calculator

2^x = e^(x * ln 2)
3^x = e^(x * ln 3)

2^x * 3^x = e^(x * ln 2 + x * ln 3) = e^(x * (ln 2 + ln 3))

so : x * (ln 2 + ln 3) = ln 8 etc.

x = 3ln(2) / [ln(3+ln(2)]

(2^x) (3^x) = 2³
x log 2 + x log 3 = 3 log 2
(x - 3) log 2 = - x log 3
x log 2 + x log 3 = 3 log 2
x (log 2 + log 3) = 3 log 2
x = 3 log 2 / (log 2 + log 3)
x = log (2 ³) / log 6
x = log 8 / log 6

2^x * 3^x = 8
6^x = 8
x ln6 = ln 8
x = (ln8) / (ln6)
x = 1.16

Good Luck!