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3^x = 5--------(equtation 1)5^y = 27------(equation 2)Let 3^x = aa = 5a^y = 273^xy = 273^xy = 3^3xy = 3
In the second equation sub 3x in place of 5Then you have 3^xy=base of 3 w/exp 3
Thanks ma'am 🙂
The answer is 3 btw this question is from which class?
Pero por qué se olvida del 5 que estaba elevado a la y? Creo que falta un paso, una explicación
I was solving it with logarythm, and it's more complicated..3^x = 53^(3log5) = 5x = (3log5)5^y = 275^(5log27) = 27y = (5log27)xy = (3log5)(5log27)xy = 3log27xy = 3
Oops, logarithmMy bad
Thank you
In the into you out 5^x, not 5^y.
By mistake,I wrote 5^x in thumbnail.
@@muqaddasadnan1743 I couldn't solve the thumbnail, so I watched the video. You tricked me ^_^
Mais tu as oublié le 5
3^x = 5--------(equtation 1)
5^y = 27------(equation 2)
Let 3^x = a
a = 5
a^y = 27
3^xy = 27
3^xy = 3^3
xy = 3
In the second equation sub 3x in place of 5
Then you have 3^xy=base of 3 w/exp 3
Thanks ma'am 🙂
The answer is 3 btw this question is from which class?
Pero por qué se olvida del 5 que estaba elevado a la y? Creo que falta un paso, una explicación
I was solving it with logarythm, and it's more complicated..
3^x = 5
3^(3log5) = 5
x = (3log5)
5^y = 27
5^(5log27) = 27
y = (5log27)
xy = (3log5)(5log27)
xy = 3log27
xy = 3
Oops, logarithm
My bad
Thank you
In the into you out 5^x, not 5^y.
By mistake,I wrote 5^x in thumbnail.
@@muqaddasadnan1743 I couldn't solve the thumbnail, so I watched the video. You tricked me ^_^
Mais tu as oublié le 5