# Question b8258

Apr 10, 2017

The question format is not clear.
Quick answer: log of a negative is not permitted so it is a no no to $x + 4 < 0$

#### Explanation:

NOT MEANT TO BE: $\left(f x\right) = {\log}_{2} \left(x + 4\right) - 3$
This would be very different!
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$\log 2$ would normally be understood as ${\log}_{10} \left(2\right)$ which is a constant.

So we have: $\text{ } f \left(x\right) = {\log}_{10} \left(2\right) \left(x + 4\right) - 3$

Set: $f \left(x\right) = y = \log \left(2\right) x + 4 \log \left(2\right) - 3$

This is the equation of a strait line

So for this condition we have:

domain (input) $\to \left(- \infty , + \infty\right)$
range (output) $\to \left(- \infty , + \infty\right)$

However; as there is no 'excluded values' in this it is a little disconcerting that mention is made of an asymptote.

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So perhaps the question is meant to be: $\left(f x\right) = {\log}_{2} \left(x + 4\right) - 3$

Set $f \left(x\right) = y = {\log}_{2} \left(x + 4\right) - 3$

Converting this to log base 10 we have

$y = {\log}_{10} \frac{x + 4}{\log} _ 10 \left(2\right) - 3$

Now this is different as you do have excluded values, in that, from ${\log}_{10} \left(x + 4\right)$

ul("the values NOT permitted are such that")" "x+4 < 0

So we have permitted value of

color(red)(x>=-4" " ->" Domain"->(-4,oo)

As ${\log}_{10} \left(2\right) < 1$ then ${\log}_{10} \frac{x + 4}{\log} _ 10 \left(2\right) > {\log}_{10} \left(x + 4\right)$

As $x$ tends to infinity then the -3 has very little effect so may be discounted

so color(green)(lim_(x->oo)" "log_10(x+4)/log_10(2)-3=oo" "->"Part of the Range"

When $x + 4$ becomes decimal then $\log \left(x + 4\right)$ becomes negative. The magnitude of the negative number increases the closer to zero $\left(x - 4\right)$ becomes. Thus :

$\textcolor{g r e e n}{{\lim}_{x + 4 \to 0} \text{ "log_10(x+4)/log_10(2)-3= -oo" "->"Part of the range}}$

color(red)("Range "(-oo,+oo)

Apr 10, 2017

See explanation.

#### Explanation:

The range is the set of all numbers for which the formula is defined.

In this example we have a ${\log}_{2}$ function which (as all logarythms) is defined only for positive values, so to calculate the domain we have to solve:

## $x + 4 > 0$

$x > - 4$

So the domain is: D=(-4;+oo)#

The ${\log}_{2}$ function takes all values, so the range is $\mathbb{R}$

The asymptote is $x = 0$, because the closer $x$ gets to zero, the smaller is $f \left(x\right)$.

${\lim}_{x \to 0} f \left(x\right) = - \infty$

The graph is:

graph{ log(x+4) - 3 [-6, 30, -17.26, 5.25]}