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
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) .
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 }
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.
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) .
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 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"
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 .
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@dennismathacademy be blessed
7ˣ + 7ˣ = 490
2*7ˣ = 2*5*7²
7ˣ = 5*7²
x = log(5)/log(7) + 2
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
Nice Alternative 👍
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) .
X= 2.8271
Use "≈", not "=" .
EXPONENCIAL
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)
👍
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 }
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.
"circumstantial"?
"I do not think it means what you think it means." -- Inigo Montoya, "The Princess Bride"
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)
Wrong! log(5/7) ≠ log(5)/log(7)
7ˣ⁻² = 5
x - 2 = log(5)/log(7)
x = log(5)/log(7) + 2
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
Completely wrong.
@5.52 how 5 comes?
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)
@prithwiraj this is clearly explained. Thanks
This is clearly explained as 7^2+log 5( base 7)
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) .
Why people donot use better dark pen??
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?
That is perfectly correct. Thank you for the comment
@slacroixfr
OK to use the caret ( ^ ) for exponents:
2^3 = 8
c^2 = c*c
@@EdLeeSB I thought of it but thought some might not understand...
@@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"
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 .