Question
1.find the zeroes of the function f(x)=log4(x+1)+log4(4x-3)
2.which of the following is the equation c^(4d+1)=7a-b written in logarithm form
2.which of the following is the equation c^(4d+1)=7a-b written in logarithm form
Answers
Reiny
log4(x+1)+log4(4x-3) = 0
log(x+1)/(log4) + log(4x+3)/log4 = 0
times log4
log(x+1) + log(4x+3) = 0
log((x+1)(4x+3)) = 0
10^0 = 4x^2 + 7x + 3
4x^2 + 7x + 2 = 0 , where x > -1
x = (√17 - 7)/8
for the 2nd, there is "no following"
log(x+1)/(log4) + log(4x+3)/log4 = 0
times log4
log(x+1) + log(4x+3) = 0
log((x+1)(4x+3)) = 0
10^0 = 4x^2 + 7x + 3
4x^2 + 7x + 2 = 0 , where x > -1
x = (√17 - 7)/8
for the 2nd, there is "no following"
Steve
log4(x+1)+log4(4x-3) = 0
log4 (x+1)(4x-3) = 0
since 4^0 = 1, we have
(x+1)(4x-3) = 1
4x^2+x-3 = 1
4x^2+x-4 = 0
x = (-1±√65)/8
check for extraneous roots, since the domain of logz is z>0
log4 (x+1)(4x-3) = 0
since 4^0 = 1, we have
(x+1)(4x-3) = 1
4x^2+x-3 = 1
4x^2+x-4 = 0
x = (-1±√65)/8
check for extraneous roots, since the domain of logz is z>0
Steve
c^(4d+1)=7a-b
log<sub><sub>c</sub></sub>(7a-b) = 4d+1
log<sub><sub>c</sub></sub>(7a-b) = 4d+1
honors student
steve is right! thank you steve!