log3^x+log7^2x=log11^4
find x?
thx
2007-05-06 03:18:26 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
That's better
So using rule 3 of logs
log3^x becomes xlog3 = 0.4771x to 4sf
log7^2x becomes 2xlog7= 1.690x to 4sf
log11^4 becomes 4log11= 4.166 to 4sf
Your equation is now
0.4771x + 1.690x = 4.166
2.1671x = 4.166
x = 1.92 to 3sf
2007-05-06 07:22:32 · answer #1 · answered by fred 5 · 0⤊ 0⤋
x log 3 + 2x log 7 = 4log 11
x = 4 log 11 / ( log 3 + 2 log 7)
2007-05-06 10:24:42 · answer #2 · answered by hustolemyname 6 · 0⤊ 0⤋
you can write
log(3^x*49^x)=log 11^4
so 147^x = 11^4
and x= log11^4/log147 =1.922
2007-05-06 10:27:38 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋
log(a) + log(b) =log(ab)
c.log(a) = log(a^c)
log3^x+log7^2x=log11^4
xlog3 + 2xlog7 = log11^4
x(log3 + 2log7)=log11^4
x(log3+log49)=log11^4
x(log147)=log(11^4)
x=4log11/log147
=1.922 (approx)
2007-05-06 10:22:30 · answer #4 · answered by gudspeling 7 · 1⤊ 0⤋
convert it as :
x log 3 + 2x log 7 = 4 log 11
(calculate all log values by u r scientific calc.)
x (0.478) + 2x (0.845) = 4.166
x [ 0.478 + 2(0.845) ] = 4.166
x = 1.922
2007-05-06 10:31:55 · answer #5 · answered by sweety 2 · 0⤊ 0⤋