Germany Olympiad Mathematics|Exponential Equation.

[@maths01n]

Good work my fellow Mathematician ❤ keep up I have subscribed

[@dennismathacademy]

Thank you Sir. I have also Subscribed to your Channel.

[@maths01n]

@dennismathacademy be blessed

[@oahuhawaii2141]

7ˣ + 7ˣ = 490 2*7ˣ = 2*5*7² 7ˣ = 5*7² x = log(5)/log(7) + 2

[@Luis-lm2lg]

EXPONENCIAL

[@АндрейПергаев-з4н]

Решение в три действия, и занимает 1 минуту, не позорьтесь 1 Действие, 7^х*(1+1)=490, 2*7^х=490 2 действие, делить на 2 7^х =245 3 действие логарифмировать по основанию 7 х=log(7)245 Всё решение Можно еще 245=5*49, тогда ответ х=2+log(7)5

[@dennismathacademy]

Nice Alternative 👍

[@oahuhawaii2141]

Everyone is so sloppy with their notation. What is log(7)5 ? Do you mean log(35) or (log(7))*5 ? If you can't write subscript 7, then use a different form: log₇(5) = log(5)/log(7) .

[@prasadrasikawidanagamachch3932]

X= 2.8271

[@oahuhawaii2141]

Use "≈", not "=" .

[@PrithwirajSen-nj6qq]

7^x + 7^x =490 > 7^x =245 x =log 245 (base 7) = log (7^2*5)(base 7) = log 7^2(base 7) + log 5(base7) = 2log 7(base 7) + log 5(base 7) = 2*1 + log 5(base7) = 2 + log 5(base 7)

[@dennismathacademy]

👍

[@oahuhawaii2141]

You should use standard notation. If you can't type in subscripts, then convert the form to divide by log(7): 7ˣ = 5*7² x = log(5*7²)/log(7) = log(5)/log(7) + 2 = log₇(5) + 2 { if you can type subscripts }

[@saladinayoubi9773]

plus astucuex en 3 lignes : 2.7^x = 7^2. 10 --> 7^(x-2) = 5 --> x-2=ln(5/7) --> x=2+ln(5/7)

[@oahuhawaii2141]

Wrong! log(5/7) ≠ log(5)/log(7) 7ˣ⁻² = 5 x - 2 = log(5)/log(7) x = log(5)/log(7) + 2

[@Alfi-rp6il]

That is much too circumstantial. You should cancel as much as possible in the very beginning: 7^x + 7^x = 490 => 2*7^2 = 2*49*5 => 7^x = 7^2 * 5 =>7^(x-2) = 5 => ln(7^(x-2)) = ln(5) => (x-2)*ln(7) = ln(5) => (x-2) = ln(5)/ln(7) => x = ln(5)/ln(7) + 2. Done.

[@oahuhawaii2141]

"circumstantial"? "I do not think it means what you think it means." -- Inigo Montoya, "The Princess Bride"

[@marceliusmartirosianas6104]

7^x+7^x 7o^70= 70*70=70^2= 7^20]=[ 2*7^x =7^2x= 7^2o 2x= 2ox=1o ACADEMIC Marcelius Martirosianas

[@oahuhawaii2141]

Completely wrong.

[@PrithwirajSen-nj6qq]

In the last line you wrote 2+ log 5/log 7 And in margin wrote log a /log b = log a (base b) But in answer you did not write 2+ log 5(base 7)

[@kamamalifestyle]

@prithwiraj this is clearly explained. Thanks

[@dennismathacademy]

This is clearly explained as 7^2+log 5( base 7)

[@oahuhawaii2141]

Everyone is so sloppy with their notation. Technically, 5 (base 7) is 5, so log 5 (base 7) is log 5 with the base of the log defaulting to 10. The use of parentheses in the proper places is required for clarity. If you can't write subscript 7, then use a different form: log₇(5) = log(5)/log(7) .

[@mjayapoornajha3832]

@5.52 how 5 comes?

[@slacroixfr]

05:34 you wrote [b to_the_power log(a/b) = b] but obviously you wanted to write [b to_the_power log(a/b) = a] because this is what you substitute a bit later, correct?

[@dennismathacademy]

That is perfectly correct. Thank you for the comment

[@EdLeeSB]

@slacroixfr OK to use the caret ( ^ ) for exponents: 2^3 = 8 c^2 = c*c

[@slacroixfr]

@@EdLeeSB I thought of it but thought some might not understand...

[@slacroixfr]

​@@dennismathacademy side note : I was actually wrong to write "log(a/b)" that means literally log of (a divided by b) although you meant logBase(b) of a Now, I do not know how to better write "logBase(b) of a" in simple text mode (with one single font size). In some programming languages we can use log(a,b) with log( expression [ , base ] ) "[ , base]" meaning "optional with default value of e"

[@oahuhawaii2141]

No divide symbol. He isn't dividing 5 by 7 and taking the log of the result. He's trying to write log base 7 of 5, and doing a confusing job by writing the 7 below 5 rather than writing "log", subscript "7", and parameter "(5)". He is sloppy in avoiding standard form: log₇(5) . You need to realize that: log₇(5) = log(5)/log(7) whereas: log(5/7) = log(5) - log(7) and they aren't equal to each other. The identity is: b^[log(a)/log(b)] = a .

[@pkgupts1153]

Why people donot use better dark pen??