2. For what value(s) of x is log (8x^2+26x-7)/log(4x-1) undefined?

You can only find the log of a positive number so

$$4x-1>0 \\
4x>1\\
x>0.25\\$$

and

$$8x^2+26x-7>0\\
(2x+7)(4x-1)>0$$

consider the parabola

y=(2x+7)(4x-1)

This is a concave up parabola with roots x=-7/2 and x=1/4  

You should sketch the parabola.

The parabola has positive y values when x>1/4 and when x<-7/2

but we already know that x must be >0.25

therefore the expression is undefined when $$x\le 0.25$$

(Are other mathematicians happy with this?)

For real numbers log(x) function isn't defined for x<=0. So for this to be undefined

$$8x^2+26x-7<0$$

$$4x-1<0$$

and further division by 0 isn't permitted so

$$\log(4x-1)= 0 \Rightarrow (4x-1)= 1$$

Sorting through this we see

$$(2x+7)(4x-1)\leq 0 \Rightarrow \left[-\dfrac 7 2 \leq x \leq \dfrac 1 4 \right]$$

This also takes care of the second condition.

We also have

$$4x-1 = 1 \Rightarrow 4x = 2 \Rightarrow x = \dfrac 1 2$$

Combing these we have that the original function is undefined on

$$\left[-\dfrac 7 2 \leq x \leq \dfrac 1 4 \right] \bigcup \left\{\dfrac 1 2 \right\}$$

Sorry Rom, but your answer would imply that the function is defined for, say, x = -5, which it isn't.

Thanks for all the answer and thanks for the LaTex Rom     

You are the best LaTex writer on the forum and now we can all benefit from your knowledge.  

If we hover over the Latex, we are given the code.  This is brilliant.  Thank you Andre.

I have added some of you output to the @@LaTex sticky page.  

I'm all excited now!!!!

Almost Melody. I think that we still have to include x = 1/2 which makes the denominator zero.

ok.

You have to put Rom's answer and my answer together to get the right answer.

For what value(s) of x is log (8x^2+26x-7)/log(4x-1) undefined?

UNDEFINED WHEN

$$\left[-\dfrac 7 2 \leq x \leq \dfrac 1 4 \right] \bigcup \left\{\dfrac 1 2 \right\}$$      (from Rom)

ALSO UNDEFINED WHEN

$$x\le 0.25$$    (from Melody)

THEREFORE undefined for

$$\left[x \leq \dfrac 1 4 \right] \bigcup \left\{\dfrac 1 2 \right\}$$

Thanks Rom and Bertie

I've thought about this problem for several days. Here's my take on the answer.

As Rom pointed out, the interval [-(7/2), (1/4)] does indeed make the "log" in the numerator undefined.

But I believe the answer to the problem lies in the denominator. In effect, we have the domain of an "inside" function -  (4x-1) - to consider, and the domain of an "outside" function - log(4x-1) - to consider. Note that any values between (1/4) and (1/2) are OK - they just make the denominator negative. But also notice that any values less than (1/4) aren't defined for log(4x-1). And x=(1/2) isn't defined, either.

So, I believe the correct answer to this problem are the two intervals (1/4, 1/2) and (1/2, ∞)

Anybody have another take??

hi Chris,

Isn't this where it IS defined?  The question asks for where it is not defined.

Yes....you are correct, Melody.......I should have said it is undefined everywhere except for my answer.......