5^a+5^a=100 .. seam 5^a= X then 2X =100 => X= 50 so 5^a= 50 => 5^a = 25+25 we can write. So 5^a = 5^2+5^2 => 5^a = 5^( 2+log2^x)+5^( 2 + log2^x) { log2^x value 1= 0} => a = 2+ log2^x Ans.. tried 👍🙂 welcome .. 30 years back done 😊 .. this math is logarithm.. tried different way calculation 😉
👍👍👍👍
😅Fantastic😅
or
5^a=100/2
5^a=50
a×log5=log50
a=log5/log5+log10/log5
a=1+1/log5
5^a=50
log5^a=log50
a log5=log(5^2×2)
a=(2 log5+log2)/log5
a=2+log5(2)
5^a+5^a=2*5^a=100
5^a=50, alog5=log50=log5+log10
A=(log5/log5=1)+(log10/log5=1/0.698=1.431)=1+1.431=2.431
5^a=50
a=log50/log5
=(2log5+log2)/log5
=2+log2/log5
5^a+5^a=100 .. seam 5^a= X then 2X =100 => X= 50 so 5^a= 50 => 5^a = 25+25 we can write. So 5^a = 5^2+5^2 => 5^a = 5^( 2+log2^x)+5^( 2 + log2^x) { log2^x value 1= 0} => a = 2+ log2^x Ans.. tried 👍🙂 welcome .. 30 years back done 😊 .. this math is logarithm.. tried different way calculation 😉
Abe jaldi solve kar kal subah panvel nikalna hai😅😅😅😂😂😂