For taking the log to base 5 as it was the appropriate log base in this case, it is better to cancel the log of base and the exponential to reduce the number of steps to solve exponentials when the log base is appropriate. The power rule is implicit when doing such a move.
^=read as to the power *=read as square root Now explain 9× *{3.(*3) =3^2 ×{3^(1/2)}×{3^(1/4)} =3^(11/4) As per question {3^(11/4)}^(1/5^x) =3^(11/4.5^x) Now 243=3^5 As per question 3^{11/4.5^x}=3^5 So, 11/4.5^x=5 So, 11/4=5^x. 5=5^(x+1) So, 5^(x+1)=11/4 Take log log{5^(x+1)}==log (11/4) (X+1).log5=log11-log4 X+1={log11-log4}/log5 X={(log11-log4 )/log5}-1..May be
Easy ❤❤
For taking the log to base 5 as it was the appropriate log base in this case, it is better to cancel the log of base and the exponential to reduce the number of steps to solve exponentials when the log base is appropriate. The power rule is implicit when doing such a move.
Thanks for sharing your expert perspective! 💯🙏🤩💕
^=read as to the power
*=read as square root
Now explain
9× *{3.(*3)
=3^2 ×{3^(1/2)}×{3^(1/4)}
=3^(11/4)
As per question
{3^(11/4)}^(1/5^x)
=3^(11/4.5^x)
Now 243=3^5
As per question
3^{11/4.5^x}=3^5
So,
11/4.5^x=5
So,
11/4=5^x. 5=5^(x+1)
So,
5^(x+1)=11/4
Take log
log{5^(x+1)}==log (11/4)
(X+1).log5=log11-log4
X+1={log11-log4}/log5
X={(log11-log4 )/log5}-1..May be
Thanks fo sharing r your in-depth knowledge 🥳🎉🎁🌲
x = approx. -0.371
(3^(11/8))^(1/(5^x))=3^5 , (11/8)^(1/(5^x))=5 , / ()^(5^x) , 11/8=5*5^x , 5^x=11/40 , x=log(11/40)/log(5) ,
test , (3^(11/8))^(1/(11/40))=3^(11/8 * 40/11) , --> 3^5=LHS , RHS --> 243=3^5 , same , OK ,
it's log (11/20) not 11/40
@@2012tulio OK , thanks ,