is there any way i can solve for x in
log (x-1) = 1/2
w/o calc ?
thanks : )
2007-11-07 15:31:22 · 7 answers · asked by Henry Chan 1 in Science & Mathematics ➔ Mathematics
raise bith sides to the power of 10:
x-1 = sqrt(10)
square both sides:
(x-1)^2 =10
x^2 -2x +1 =10
x^2 -2x -9 = 0
x=[2 +/- sqrt(40) ] 2
2007-11-07 15:38:39 · answer #1 · answered by stevemorris1 5 · 0⤊ 0⤋
Assuming log means natural log,
x - 1 = e^(1/2)
x= 1 + âe.
2007-11-07 15:40:41 · answer #2 · answered by steiner1745 7 · 0⤊ 1⤋
Assuming the question is as written with base 10
rewrite as an exponential equation
10^.5=x-1
x= 1 + square root of 10
2007-11-07 15:35:05 · answer #3 · answered by imalava 2 · 0⤊ 0⤋
10^log(x-1) = 10^(1/2)
x-1 = 3.162
x = 4.162
2007-11-07 15:34:42 · answer #4 · answered by robert 6 · 0⤊ 0⤋
x-1 = sqrt(10)
x = 1 + sqrt(10)
there a some ways for evaluating square roots manually if you need a decimal approximation. but it would be tedious to try to explain here.
2007-11-07 15:36:12 · answer #5 · answered by Anonymous · 0⤊ 0⤋
i dunno, but your the 4th Henry Chan ive known in my life lol
maybe it has something to do with ln natural logs i dono i forgot that stuff sorry.
2007-11-07 15:37:01 · answer #6 · answered by Anonymous · 0⤊ 1⤋
idk
2007-11-07 15:34:53 · answer #7 · answered by Anonymous · 0⤊ 1⤋