Ok, pretty nice. Not gonna lie the 2^10 + 2^10 being 2*2^10 didn't need that verbose of an explanation but sure. Also in the first equation I would write that in the "Rejected" case it just has no Real solutions so that's why we "reject" it. And also in the final formula x = log_2((1+√5)/2) I'd simplify with the rule that log(x/y) = log(x) - log(y), so x = log_2(1+√5) - log_2(2) = log_2(1+√5) - 1.
(X)^9+((x)^6=36 {(X)^3}^3+{(x)^3}^2=36 Let x^3=a So, a^3+a^2-36=0 In the cubic equation the first root will be 3 by hit & trail method a=3 So, X^3=3 X=(3)^(1/3) For other roots (a-3)(a^2+4a+12)=0 a^2+4a+12=0 a={(-4±4.*2i)}/2 X=[{(-4±4.*2i)}/2]^(1/3)
respectfully, your first solution is wrong. it's log_2(1+sqrt5)-1. Your solution can be simplified (which it should be) but you still made a mistake with the sign
How can anyone be so idiotic to make such an easy task so complicated? 100^x=(10^x)^2 With that, you get a quadratic equation that can easily be solved. Calculating the log of the solution gives you the final answer: X=log phi phi=1.618033… The golden ratio
Too complicated. 1^x is always 1 so 1 + 10^x = 100^x. Rearrange. 100^x - 10^x = 1. 100^x is the same as 10^(2x). 10^2x - 10^x - 1 = 0 Let y = 10^x. y^2 - y -1 = 0. Use quadratic formula. y = 1 +/- sqrt(5) / 2. 10^x = 1+sqrt(5) / 2 x = log(1 + sqrt(5)/2) / log(10)
The very first step of your solution is incorrect. (1^x + 10^x)/100^x is not equal to 1^x/100^x + 10^x/100^x. Also you can easily see that 1^4 + 10^4 does not equal 100^4. The actual answer you get by knowing that 1^x = 1 for all x and by then using u substitution on 10^x and solving that quadratic
In the initial division, it needs to be pointed out 125^x cannot equal 0, so that it is okay to do this division. Also, consider immediately letting u = 5^x. Then u + u^2 = u^3, 0 = u^3 - u^2 - u = u(u^2 - u - 1). 5^x = 0 has no solutions. We get to the same situation, with u^2 - u - 1 = 0 much more quickly. (I used u instead of y.)
