# Find the range of logarithm function [duplicate]

Suppose there is a function $$f(x)=\log_3(5+4x-x^2)$$ Then find the range of $$f(x)$$.

I don't know how to find range of functions like $$g(x)=\log_n(h(x))$$ But still I decided to give it a try.

Since $$f(x)$$ is a log function, we have $$x\in(-1,5)$$ Let us denote $$5+4x-x^2=j(x)$$

So the minimum value of $$j(x)$$ in the interval $$x\in(-1,5)$$ is $$\lim j(x)\rightarrow0$$ and the maximum value is $$\lim j(x)\rightarrow5$$. After this I am stuck.

I actually want to know how to find range of a function like this $$b(x)=\log_n(ax^2+bx+c)$$

Any help is greatly appreciated.

• Well, in fact, here you go. Commented Sep 6, 2022 at 12:20

$$5+4x-x^2=-(x^2-4x+4)+9=-(x-2)^2+9=(3-(x-2))(3+(x-2))=-(x+1)(x-5)$$

Sine $$-(x+1)(x-5)>0$$ when $$x\in(1,5)$$, the function is defined when $$x\in(1,5)$$.

On this interval $$f(x)\to-\infty$$ as $$x\to -1$$ and $$f(x)\to-\infty$$ as $$x\to 5$$.

Next, $$f'(x)=\frac{4-2x}{\ln(3)(5+4x-x^2)}$$.

$$f'(x)=0$$, when $$x=2$$.

At this point there is a maximum, thus the range of the function is $$(-\infty, f(2)]=(-\infty, \log_3(9)]$$.

Logarithm to any base is a strictly increasing continuous function. The range of any continuous fucntion on $$\mathbb R$$ or an interval in $$\mathbb R$$ is an interval and so the range of $$\log_n (f(x))$$ is simply the interval from $$f(a)$$ to $$f(b)$$ where $$a,b$$ are the end points of the arnge of $$f$$. I am ignoring the end points here but you should be able to see whether the end poinst are included or not in the present case. [The answer in this case is $$(-\infty, 2]$$].

There is no minimum for $$f$$. There is a maximum somewhere, and you need to find it. Say the maximal value is $$m$$, then the range is $$(-\infty,m]$$. In fact, it is easy to see that the maximum is attained at $$x=2$$, so that the maximal value is $$m=\log_3(9)=2$$, hence the range is $$(-\infty,2]$$.