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.
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.
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...
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.
Please share your resolution
@@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).
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.
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.
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...