lecture 4 exponential and logarithmic functions part 2 ♡

To evaluate $lo g_{a} (b)$ , we must restrict $b \in (0, \infty)$ and $a \in (0, \infty) \ {1}$ .

Need to memorize some log identities, see logarithm.

Let $f : (0, \infty) \to R$ , where $f (x) = ln (x)$ and $g : R \to R$ , where $g (x) = x^{2} + 1$ .

Find the rule for h where $h (x) = f (g (x))$ . State the domain and range of h.

Function g has domain R, and it outputs a positive number, which fits in the domain of f, Therefore, function h has domain R.

dom h = R

The rule is $h (x) = ln (x^{2} + 1)$ .

The range is the same as the range of f.

We know the value that g(x) inputs to f(x), $x^{2} + 1$ , is always greater than 1.

We know that the graph of $ln (x)$ has a x-intercept at (1,0).

Therefore

ran h = $[0, \infty)$

Show that $h (x) + h (- x) = f ((g (x))^{2})$
$h (x) + h (- x) = ln (x^{2} + 1) + ln (- x)^{2} + 1) = ln ((x^{2} + 1)^{2}) = f (g (x))^{2})$

Let $k : (- \infty, 0] \to R$ , where $k (x) = ln (x^{2} + 1)$ .

Find the rule for $k^{- 1}$ .
Let $y = k^{- 1} (x)$

$k (x) = ln (x^{2} + 1)$
$x = ln (y^{2} + 1)$
$e^{x} = y^{2} + 1$
$y^{2} = e^{x} - 1$
$y = \pm e^{x} - 1$

$k^{- 1} (x) = \pm e^{x} - 1$

State the domain and range of $k^{- 1}$ .

Let $h : R \to R, h (t) = e^{- t} (t^{2} + a t + 10)$ , where a is a real number. Given $h^{'} (t) = e^{- t} (- t^{2} + (2 - a) t + a - 10)$ ,

Find values of a such that the graph $y = h (t)$ has exactly one stationary point.
At a stationary point, h’(t) = 0.

$0 = e^{- t} \cdot (- t^{2} + (2 - a) t + a - 10)$
Given this equation, we know at least one of $e^{- t}$ or $(- t^{2} + (2 - a) t + a - 10)$ must be zero. Since $e^{- t}$ is always positive, therefore
$0 = - t^{2} + (2 - a) t + a - 10$
We are looking for cases where $y = h (t)$ has exactly one stationary point. This means the discriminant is zero.
$Δ = (2 - a)^{2} - 4 (- 1) (a - 10)$
$Δ = 4 - 4 a + a^{2} + 4 a - 40$
$Δ =$

Find values of a such that $h^{'} (t) < 0$ for all real values of t.

Find the range of the function $f (x) = \frac{a ^{x} - 4}{a ^{x} - 2}$ where $a > 0$ and $a \neq = 1$ .

To find the range, we should graph it.

The structure of this function is new, so we will need to graph it using deductions.

We see that the function has a restriction that $a^{x} - 2 \neq = 0$ , which means $x \neq = lo g_{a} (2)$ . Therefore, there is a vertical asymptote $x \neq = lo g_{a} (2)$ .

Solve $\frac{1}{2} ln (9 x^{2} - 30 x + 25) - \frac{1}{4} ln (81) = 0$ .

$\frac{1}{2} ln ((3 x - 5)^{2}) - ln (8 1^{\frac{1}{4}}) = 0$

Notice the restriction $9 x^{2} - 30 x + 25 \neq = 0$ which is $3 x - 5 \neq = 0$ which is $x \neq = \frac{5}{3}$ .

$ln (∣3 x - 5∣) - ln (3) = 0$

$ln (\frac{∣3 x - 5∣}{3}) = 0$

$ln (∣ x - \frac{5}{3} ∣) = ln (1)$

$∣ x - \frac{5}{3} ∣ = 1$
$x - \frac{5}{3} = \pm 1$
$x = \frac{2}{3} \lor x = \frac{8}{3}$

Restrictions satisfied.

Umm we got another way to do this.

$\frac{1}{2} ln (9 x^{2} - 30 x + 25) - \frac{1}{4} ln (81) = 0$

$ln ((9 x^{2} - 30 x + 25)^{\frac{1}{2}}) = ln (8 1^{\frac{1}{4}})$

Notice the restriction $9 x^{2} - 30 x + 25 \neq = 0$ which is $3 x - 5 \neq = 0$ which is $x \neq = \frac{5}{3}$ .

$(9 x^{2} - 30 x + 25)^{\frac{1}{2}} = 8 1^{\frac{1}{4}}$

$(9 x^{2} - 30 x + 25)^{\frac{1}{2}} = 3$

$9 x^{2} - 30 x + 25 = 9$

$9 x^{2} - 30 x + 16 = 0$

$x = \frac{30 \pm 18}{18}$

$x = \frac{2}{3} \lor x = \frac{8}{3}$

The graph of $f (x) = a e^{x} + b$ passes through the points $(0, - 10)$ and $(ln (2), - 17)$ . Find the values of a and b.

$- 10 = a + b$
$- 17 = a e^{l n (2)} + b$

$b = - 10 - a$

$- 17 = a e^{l n (2)} - 10 - a$
$0 = 2 a - 10 - a + 17$
$0 = a + 7$
$a = - 7$
$b = - 3$

The graph of $f (x) = e^{a x + b} - 2$ passes through the points $(1, 2 e - 2)$ and $(0, e - 2)$ . Find the values of a and b.

$2 e - 2 = e^{a + b} - 2$

$e - 2 = e^{b} - 2$

$e = e^{b}$
$b = 1$

$2 e - 2 = e^{a + 1} - 2$
$2 e = e^{a + 1}$

$2 = e^{a}$
$a = ln (2)$

Given $2^{a} = 5^{b} = 10$ , evaluate $\frac{1}{a} + \frac{1}{b}$ .

$a = lo g_{2} (10)$
$b = lo g_{5} (10)$
$\frac{1}{a} + \frac{1}{b} = lo g_{10} (2) + lo g_{10} (5) = 1$

Find the range of $y = f (x)$ for $f : (- 2, 0) \to R, f (x) = 4 - 9 e^{- 4 x - 8}$ .

We need to graph it.

$f (x) = - 9 e^{- 4 (x + 2)} + 4$

Restrictions are:

Dilation by 9 in x axis.
Reflection in x axis.
Dilation by 4 in y axis.
Reflection in y axis.
Translate 2 left.
Translate 4 up.

We know what the graph roughly looks like.
We know the asymptote is $y = 4$ .

Range can be $(- \infty, 4])$ .

Add domain restrictions now.

$f (- 2) = - 5$
$f (0) = 4 - 9 e^{- 8}$

Therefore the range is $(- 5, 4 - 9 e^{- 8})$

Simplify $\frac{e ^{- x} \cdot 8 ^{x} \cdot ( e ^{4 x} + e ^{x} )}{16 ( e ^{3 x} + 1 )}$ .

$= \frac{8 ^{x} \cdot ( e ^{3 x} + 1 )}{16 ( e ^{3 x} + 1 )}$
$= \frac{8 ^{x}}{16}$
$= \frac{2 ^{3 x}}{2 ^{4}}$
$= 2^{3 x - 4}$

Let $f : R \to R, f (x) = 3 e^{x} + 2$ a\in(0,\infty) \backslash {1}and let $g : (1, \infty) \to R, g (x) = 2 ln ( x)$ . Find $f (g (x))$ in the form $a x^{2} + b$ .

$f (g (x)) = 3 e^{2 l n (x)} + 2$ | $x \in (1, \infty)$

$f (g (x)) = 3 e^{l n (x)^{2}} + 2$ | $x \in (1, \infty)$

$f (g (x)) = 3 x^{2} + 2$ | $x \in (1, \infty)$

The graph $y = - 5 e^{a x + b} + 5$ passes through points $(0, 5 - 5 e)$ and $(- \frac{1}{3}, 0)$ . Find the values of a and b.

$5 - 5 e = - 5 e^{b} + 5$
$0 = - 5 e^{\frac{- a}{3} + b} + 5$

$- 5 e = - 5 e^{b}$
$e = e^{b}$
$b = 1$
$0 = - 5 e^{\frac{- a}{3} + 1} + 5$
$0 = e^{\frac{- a + 3}{3}} - 1$
$1 = e^{\frac{- a + 3}{3}}$
$e^{0} = e^{\frac{- a + 3}{3}}$
$0 = - a + 3$
$a = 3$

Solve $22 \cdot 3^{2 x} + 3^{x + 2} - 1 = 0$ .

$22 \cdot 3^{2 x} + 9 \cdot 3^{x} - 1 = 0$
Let $k = 3^{x}$
$22 k^{2} + 9 k - 1 = 0$
$(11 k - 1) (2 k + 1) = 0$
$k = \frac{1}{11} \lor k = \frac{- 1}{2}$

$3^{x} = \frac{1}{11} \lor 3^{x} = \frac{- 1}{2}$
$x = lo g_{3} (\frac{1}{11}) \lor x = lo g_{3} (\frac{- 1}{2})$

Second impossible.
$x = lo g_{3} (\frac{1}{11})$
$x = - lo g_{3} (11)$

Given $u = lo g_{9} (x)$ and $3^{6} = 729$ , express $lo g_{x} (729)$ in terms of u.

$lo g_{x} (729) = \frac{lo g _{9} ( 729 )}{lo g _{9} ( x )}$
$= \frac{3}{lo g _{9} ( x )}$
$= \frac{3}{u}$

Solve $lo g_{x} (100) + lo g_{x^{2}} (100) = 3$ .

$lo g_{x} (100) + \frac{lo g _{x} ( 100 )}{lo g _{x} ( x ^{2} )} = 3$
$lo g_{x} (100) + \frac{lo g _{x} ( 100 )}{lo g _{x} ( x ^{2} )} = 3$

$lo g_{x} (100) + \frac{lo g _{x} ( 100 )}{2} = 3$
$\frac{3}{2} lo g_{x} (100) = 3$
$lo g_{x} (100) = 2$
$x = 100$
$x = \pm 10$

To evaluate $lo g_{a} (b)$ , we must restrict $b \in (0, \infty)$ and $a \in (0, \infty) \ {1}$ .
$x = 10$

Solve $35 e^{2 x} - e^{x} - 12 = 0$ .

Let $k = e^{x}$
$35 k^{2} - k - 12 = 0$
$Δ = 1 - 4 (35 \cdot - 12)$
$Δ = 1681$
$k = \frac{1 \pm 1681}{70}$
$e^{x} = \frac{1 \pm 41}{70}$
$x = ln (\frac{1 \pm 41}{70})$
$\frac{1 \pm 41}{70} > 0$
$x = ln (\frac{42}{70})$
$x = ln (\frac{3}{5})$

Find a and b where $y = ln (a x + b) - 4$ given it passes through $(0, - 2)$ and the asymptote is $x = \frac{e ^{2}}{3}$ .

$- 2 = ln (b) - 4$
$2 = ln (b)$
$e^{2} = b$

$\frac{a e ^{2}}{3} + b = 0$
$\frac{a e ^{2}}{3} + e^{2} = 0$
$a e^{2} + 3 e^{2} = 0$
$(a + 3) e^{2} = 0$
$a = - 3$

Find a and b where $y = ln (a x + b) + 6$ given it passes through $(2, 6)$ and the asymptote is $x = \frac{7}{4}$ .

$6 = ln (2 a + b) + 6$
$ln (2 a + b) = 0$
$2 a + b = 1$ | $2 a + b > 0$
$2 a + b = 1$

$\frac{7}{4} a + b = 0$
$7 a + 4 b = 0$

$8 a + 4 b = 4$
$a = 4$
$28 + 4 b = 0$
$b = - 7$

Given that the graph of the function $f (x) = 2 lo g_{4} (a x + b) - 3$ passes through the points $(0, - 3)$ and $(1, 1)$ . Find the values of a and b.

$- 3 = 2 lo g_{4} (b) - 3$
$1 = 2 lo g_{4} (a + b) - 3$

$2 lo g_{4} (b) = 0$
$lo g_{4} (b) = 0$
$b = 1$

$1 = 2 lo g_{4} (a + 1) - 3$
$4 = 2 lo g_{4} (a + 1)$
$2 = lo g_{4} (a + 1)$
$16 = a + 1$
$a = 15$

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lecture 4 exponential and logarithmic functions part 2

Let $f : (0, \infty) \to R$ , where $f (x) = ln (x)$ and $g : R \to R$ , where $g (x) = x^{2} + 1$ .

Let $k : (- \infty, 0] \to R$ , where $k (x) = ln (x^{2} + 1)$ .

Let $h : R \to R, h (t) = e^{- t} (t^{2} + a t + 10)$ , where a is a real number. Given $h^{'} (t) = e^{- t} (- t^{2} + (2 - a) t + a - 10)$ ,

Find the range of the function $f (x) = \frac{a ^{x} - 4}{a ^{x} - 2}$ where $a > 0$ and $a \neq = 1$ .

Solve $\frac{1}{2} ln (9 x^{2} - 30 x + 25) - \frac{1}{4} ln (81) = 0$ .

The graph of $f (x) = a e^{x} + b$ passes through the points $(0, - 10)$ and $(ln (2), - 17)$ . Find the values of a and b.

The graph of $f (x) = e^{a x + b} - 2$ passes through the points $(1, 2 e - 2)$ and $(0, e - 2)$ . Find the values of a and b.

Given $2^{a} = 5^{b} = 10$ , evaluate $\frac{1}{a} + \frac{1}{b}$ .

Find the range of $y = f (x)$ for $f : (- 2, 0) \to R, f (x) = 4 - 9 e^{- 4 x - 8}$ .

Simplify $\frac{e ^{- x} \cdot 8 ^{x} \cdot ( e ^{4 x} + e ^{x} )}{16 ( e ^{3 x} + 1 )}$ .

Let $f : R \to R, f (x) = 3 e^{x} + 2$ a\in(0,\infty) \backslash {1}and let $g : (1, \infty) \to R, g (x) = 2 ln ( x)$ . Find $f (g (x))$ in the form $a x^{2} + b$ .

The graph $y = - 5 e^{a x + b} + 5$ passes through points $(0, 5 - 5 e)$ and $(- \frac{1}{3}, 0)$ . Find the values of a and b.

Solve $22 \cdot 3^{2 x} + 3^{x + 2} - 1 = 0$ .

Given $u = lo g_{9} (x)$ and $3^{6} = 729$ , express $lo g_{x} (729)$ in terms of u.

Solve $lo g_{x} (100) + lo g_{x^{2}} (100) = 3$ .

Solve $35 e^{2 x} - e^{x} - 12 = 0$ .

Find a and b where $y = ln (a x + b) - 4$ given it passes through $(0, - 2)$ and the asymptote is $x = \frac{e ^{2}}{3}$ .

Find a and b where $y = ln (a x + b) + 6$ given it passes through $(2, 6)$ and the asymptote is $x = \frac{7}{4}$ .

Given that the graph of the function $f (x) = 2 lo g_{4} (a x + b) - 3$ passes through the points $(0, - 3)$ and $(1, 1)$ . Find the values of a and b.

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lecture 4 exponential and logarithmic functions part 2

Let f:(0,∞)→R, where f(x)=ln(x) and g:R→R, where g(x)=x2+1.

Let k:(−∞,0]→R, where k(x)=ln(x2+1).

Let h:R→R,h(t)=e−t(t2+at+10), where a is a real number. Given h′(t)=e−t(−t2+(2−a)t+a−10),

Find the range of the function f(x)=ax−2ax−4​ where a>0 and a=1.

Solve 21​ln(9x2−30x+25)−41​ln(81)=0.

The graph of f(x)=aex+b passes through the points (0,−10) and (ln(2),−17). Find the values of a and b.

The graph of f(x)=eax+b−2 passes through the points (1,2e−2) and (0,e−2). Find the values of a and b.

Given 2a=5b=10, evaluate a1​+b1​.

Find the range of y=f(x) for f:(−2,0)→R,f(x)=4−9e−4x−8.

Simplify 16(e3x+1)e−x⋅8x⋅(e4x+ex)​.

Let f:R→R,f(x)=3ex+2 a\in(0,\infty) \backslash {1}and let g:(1,∞)→R,g(x)=2ln(​x). Find f(g(x)) in the form ax2+b.

The graph y=−5eax+b+5 passes through points (0,5−5e) and (−31​,0). Find the values of a and b.

Solve 22⋅32x+3x+2−1=0.

Given u=log9​(x) and 36=729, express logx​(729) in terms of u.

Solve logx​(100)+logx2​(100)=3.

Solve 35e2x−ex−12=0.

Find a and b where y=ln(ax+b)−4 given it passes through (0,−2) and the asymptote is x=3e2​.

Find a and b where y=ln(ax+b)+6 given it passes through (2,6) and the asymptote is x=47​.

Given that the graph of the function f(x)=2log4​​(ax+b)−3 passes through the points (0,−3) and (1,1). Find the values of a and b.

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Let $f : (0, \infty) \to R$ , where $f (x) = ln (x)$ and $g : R \to R$ , where $g (x) = x^{2} + 1$ .

Let $k : (- \infty, 0] \to R$ , where $k (x) = ln (x^{2} + 1)$ .

Let $h : R \to R, h (t) = e^{- t} (t^{2} + a t + 10)$ , where a is a real number. Given $h^{'} (t) = e^{- t} (- t^{2} + (2 - a) t + a - 10)$ ,

Find the range of the function $f (x) = \frac{a ^{x} - 4}{a ^{x} - 2}$ where $a > 0$ and $a \neq = 1$ .

Solve $\frac{1}{2} ln (9 x^{2} - 30 x + 25) - \frac{1}{4} ln (81) = 0$ .

The graph of $f (x) = a e^{x} + b$ passes through the points $(0, - 10)$ and $(ln (2), - 17)$ . Find the values of a and b.

The graph of $f (x) = e^{a x + b} - 2$ passes through the points $(1, 2 e - 2)$ and $(0, e - 2)$ . Find the values of a and b.

Given $2^{a} = 5^{b} = 10$ , evaluate $\frac{1}{a} + \frac{1}{b}$ .

Find the range of $y = f (x)$ for $f : (- 2, 0) \to R, f (x) = 4 - 9 e^{- 4 x - 8}$ .

Simplify $\frac{e ^{- x} \cdot 8 ^{x} \cdot ( e ^{4 x} + e ^{x} )}{16 ( e ^{3 x} + 1 )}$ .

Let $f : R \to R, f (x) = 3 e^{x} + 2$ a\in(0,\infty) \backslash {1}and let $g : (1, \infty) \to R, g (x) = 2 ln ( x)$ . Find $f (g (x))$ in the form $a x^{2} + b$ .

The graph $y = - 5 e^{a x + b} + 5$ passes through points $(0, 5 - 5 e)$ and $(- \frac{1}{3}, 0)$ . Find the values of a and b.

Solve $22 \cdot 3^{2 x} + 3^{x + 2} - 1 = 0$ .

Given $u = lo g_{9} (x)$ and $3^{6} = 729$ , express $lo g_{x} (729)$ in terms of u.

Solve $lo g_{x} (100) + lo g_{x^{2}} (100) = 3$ .

Solve $35 e^{2 x} - e^{x} - 12 = 0$ .

Find a and b where $y = ln (a x + b) - 4$ given it passes through $(0, - 2)$ and the asymptote is $x = \frac{e ^{2}}{3}$ .

Find a and b where $y = ln (a x + b) + 6$ given it passes through $(2, 6)$ and the asymptote is $x = \frac{7}{4}$ .

Given that the graph of the function $f (x) = 2 lo g_{4} (a x + b) - 3$ passes through the points $(0, - 3)$ and $(1, 1)$ . Find the values of a and b.