To solve exponential equations where the variable appears in the exponent, apply logarithms to both sides and use logarithm properties such as log(m^n) = n*log(m), log(mn) = log(m) + log(n), and log(m/n) = log(m) - log(n) to simplify and isolate the variable. For the equation 3^a * 3^a * 3^a = 36, this yields a = 2 + (2*log(2)/log(3))/3.
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Solving a 'Stanford' University entrance exam | a=?Hinzugefügt:
Hello Friends find the value of 'a' If 3^a.3^a.3^a=36 let's have a solution as we know x.x.x=x^3 so It will be (3^a)^3=36 as (a^m)^n=a^mn we have 3^3a=36 take 'log' on both sides log3^3a=log36 as logm^n=nlogm 3alog3=log36 divide by 'log3' on both sides 3alog3/log3=log36/log3 log3 and log3 cancels 3a=log36/log3 as 36=9.4 3a=log(9.4)/log3 logmn=logm+logn 3a=log9+log4/log3 3a=log9/log3+log4/log3 as 4=2^2 and 9=3^2 3a=log3^2/log3+log2^2/log3 3a=2log3/log3+2log2/log3 log3 and log3 cancels 3a=2+2log2/log3 as loga/logb=loga(b) 3a=2+2log2(3) divide by '3' on both sides 3a/3=2+2log2(3)/3 3 and 3 cancels a=2+2log2(3)/3 the value of 'a' now verify 3^a.3^a.3^a=36 3^2+2log2(3).3^2+2log2(3).3^2+2log2(3)=36 base same, powers add 3^3(2+2log2(3)/3)=36 3 and 3 cancels 3^2+2log2(3)=36 a^m+n=a^m.a^n 3^2.3^2log2(3)=36 as nlogm=logm^n we have 9.3^log2^2(3)=36 9.3^log4(3)=36 as we know a^logb(a)=b It will be 9.4=36 36=36 which shows that a=2+2log2(3)/3 satisfies the equation as given above thanks for watching this video please subscribe this channel to get the notification of my new videos and don't forget to share these videos with your classmates and friends so that they also have a benefit of it you can also visit the Playlists of this channel to learn more and more ok bye
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