USA Olympiad Exponential Equation: find k!

[@keeteo82]

I thought you were going to find k! (k factorial) :) . You could divide both sides of 5^x=50 by 25=5^2 to simplify the problem to 5^(x-2)=2. It would be easier to apply log base 5

[@wigpiipgiw1582]

How did you jump to 5^k =2.25?

[@BZKnowHow]

I did not jump directly to 5^k=2.25..... .I have solved it completely following algebra rules...please watch again Sir

[@wigpiipgiw1582]

@@BZKnowHow sorry, I'm British and we use different notation so didn't understand what you were in that step, after goofling I understand it now

[@xchanciex]

2*25

[@oahuhawaii2141]

2*5ᵏ = 100 5ᵏ = 50 k = log(50)/log(5) k = (1 + log(5))/log(5) k = 1/log(5) + 1 = 1/(1 - log(2)) + 1 ≈ 2.4306765580733... Estimate: k ≈ 1/(1 - .30103000) + 1 ≈ 2.4306765 669485... Accuracy is 8 sig figs given the log(2) estimate.

[@BZKnowHow]

thats also right

[@purpleblue17]

K is 10 .... 50 + 50 = 100

[@oahuhawaii2141]

Why do you write log₅2 as log₅² ? It's confusing, as I'm expecting log₅x² .

[@BZKnowHow]

dear ...it was log 25 so i wrote it as log5^2 and as per rule we will bring exponent to the left side of log so it became 2 log5. I hope its clear now.