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Finally.. 3^ logx= 6 only valid soln. Hence x ^ log 3= 6 OrX=(6)^1/ log3= approx 43.1 soln.
Let y=x^log3. Then, log x = logy/log3. So, 27^[logy/log3 -1] = y+2 > [logy/log3 -1] log 27 = log(y+2) > 3[log y - log 3] = log(y+2) > 3 log(y/3) = log(y+2) > log(y^3/27) = log(y+_2) > y^3=27y+54 > y= -3,-3,6. But y cannot be negative for real x. Thus, x^log3 = 6 > x= log6/log3. So, x= 10^(1+log2/log3).
27^(logx-1)-x^log3=227^(logx)/27-3^logx=2(3^logx)^3/27-3^logx=2Let 3^logx=a ≥ 0a³/27-a=2a³-27a-54=0(a-6)(a+3)²=0 hence a=6 (a=-3 rejected as negative)3^logx=6 log both sides(logx)(log3)=log6logx=log6/log3=log(6^(1/log3))x=6^(1/log3)
Ισχυει 3^(logχ)=χ^(log3) χ>0 θετω3^(logχ)=y>0....καταληγω y^3-27y-54=0 ; (y-6)(y^2+6y+9)=0 ; y=6 ή y=-3 απορριπτεται. Αρα 3^(logχ)=6 ; logχ×log3=log6 ; logχ=(log6)/(log3) ; χ=(10)^[log6/log3]
Finally.. 3^ logx= 6 only valid soln. Hence x ^ log 3= 6 Or
X=(6)^1/ log3= approx 43.1 soln.
Let y=x^log3. Then, log x = logy/log3. So, 27^[logy/log3 -1] = y+2 > [logy/log3 -1] log 27 = log(y+2) > 3[log y - log 3] = log(y+2) > 3 log(y/3) = log(y+2) > log(y^3/27) = log(y+_2) > y^3=27y+54 > y= -3,-3,6. But y cannot be negative for real x. Thus, x^log3 = 6 > x= log6/log3. So, x= 10^(1+log2/log3).
27^(logx-1)-x^log3=2
27^(logx)/27-3^logx=2
(3^logx)^3/27-3^logx=2
Let 3^logx=a ≥ 0
a³/27-a=2
a³-27a-54=0
(a-6)(a+3)²=0 hence a=6 (a=-3 rejected as negative)
3^logx=6 log both sides
(logx)(log3)=log6
logx=log6/log3=log(6^(1/log3))
x=6^(1/log3)
Ισχυει 3^(logχ)=χ^(log3) χ>0 θετω3^(logχ)=y>0....καταληγω y^3-27y-54=0 ; (y-6)(y^2+6y+9)=0 ; y=6 ή y=-3 απορριπτεται. Αρα 3^(logχ)=6 ; logχ×log3=log6 ; logχ=(log6)/(log3) ; χ=(10)^[log6/log3]