Question

Why is there no x-intercept for this equation? How do I show it algebraically?
y=log4(4x+8)+1

Answer 1

\(x\) intercept mean where \(y=0\) set \[\log_4(4x+8)+1=0\] and solve for \(x\)

Answer 2

Yeah I get 4^-2.5

Answer 3

frankly i don't ee what that cannot be done

Answer 4

The problem is that I keep getting an answer when the teacher said there isn't one

Answer 5

log(A) is not defined for A <= 0

Answer 6

\[\log_4(4x+8)+1=0\\
\log_4(4x+8)=-1\\
4x+8=4^{-1}=\frac{1}{4}\] why not? look like you can solve it to me

Answer 7

i get \(x=-\frac{31}{16}\)

Answer 8

which is not the answer you got, but it is an answer
your math teacher i wrong, this crosses the \(x\) axis for sure

Answer 9

thank you

Answer 10

yw

Answer 11

@ranga I am with you , however, in this case A >0

Answer 12

?

Answer 13

I did not do the calculation. I was looking at one possible reason why this function may not have a root.

Answer 14

you cannot take the log of a negative number which means to the domain of this function is \(x\geq -2\)

Answer 15

4(-31/16 )+8 >0

Answer 16

but the range of a logarithmic function is \((-\infty, \infty)\)

Answer 17

Its undefined because the curve approaches the y axis but never touches it.

Answer 18

Domain is x>0

Answer 19

In this case the domain is > -2. (-2, infinity)

Answer 20

That agrees with @ranga 's statement.