An Interesting Exponential Equation

[@Blaqjaqshellaq]

If you include complex solutions, there's also x=[ln(phi)+i*(2*n+1)*pi]/[ln(2)-ln(3)], phi being the golden ratio and n being any integer.

[@Wrest_1349]

Please share your resolution

[@Blaqjaqshellaq]

@@Wrest_1349 e^x can also equal -phi/[ln(2)-ln(3)], ln(-1)=i*(2*n+1) and ln(-phi)=ln(phi) + ln(-1).

[@1234larry1]

Seems like if you intended to check the solution into the equation, then using common log would be easier than the natural log, since you can change the base.

[@Foamea45]

You overcomplicated this problem.The first thing was right,you divide by 9^x but you write (4/9)^x as [(2/3)^x]^2 and substitute 2/3 as let's say a,but a>0 because an exponential is always positive.Then you have a quadratic in the variable you chose(for me it is a^2+a-1=0) and the solutions are(-1±5^0.5)/2 and since a^0,5 is approximately 2.23 and a is strictly positive we wil chose the +delta root. then we write a as (2/3)^x and x is log in base 2/3 of (-1+5^0,5)/2 ,the same answer,shorter path.

[@zawatsky]

3^2x-2^2x-3^x*2^x=0. 3^x:=a, 2^x:=b, a²-2ab-b²=0. (a+b)²=2b²⇔a+b=√2b, но b=2^x⇒√2b=2^(x+½). 2^x+3^x=2^x*(1+(½)^x)=2^(x+lb(1+(½)^x)).⇔x+½=x+lb(1+(½)^x)⇔lb(1+(½)^x)=½⇔√2=1+(½)^x⇔(½)^x=√2-1=1/(2^x)⇔2^x=1/(√2-1)⇔2^(-x)=√2-1⇔-x=lb(√2-1)⇔x=-lb(√2-1). Жаль, что не просто √2, но наверняка можно упростить ещё...