Exponential and also Logarithmic Functions

Part 1. Exponential features

The exponential function v base $b$ is identified for $x in chathamtownfc.netbbR$ through $f(x) = b^x,$ where $b > 0$ and $b eq 1.$Graphically, we can see the complying with properties:

Furthermore, because $b > 0$ (by definition), we know that $f(x) = b^x eq 0$ for any type of $x in chathamtownfc.netbbR.$ Consequently, we have a horizontal asymptote in ~ the line $y=0.$
The above properties the increasing and decreasing show that exponential features are $1-1,$ and therefore have actually inverses (which will be discussed in part 2).The natural exponential function is recognized as $y=e^x,$ whereby $e$ is Euler’sirrational number: $2.71828cdots$

Example. resolve the adhering to exponential equations:

1. $2^x4^3x+1 = frac2sqrt8 $
Solution: One strategy is come express both political parties in terms of the very same base, namely $b=2,$ so the the properties of exponents can be used.eginalign*& 2^x (2^2)^3x+1 = frac2(2^3)^1/2 \& Rightarrow 2^x+6x+2 = 2^1-(3/2) \& Rightarrow 2^7x+2 = 2^-(1/2) endalign*Now, we have actually that $f(7x+2) = fleft( -frac12 ight),$ wherein $f(x) = 2^x,$ and also because exponential features are $1-1,$ we can conclude the $7x+2 = -frac12.$Therefore, $x = -frac514.$
2. $ 3x(e^x) + x^2 (e^x) = 0 $
Solution. First notice that $xcdot e^x$ is a typical factor. eginalign* (x)(e^x)(3+x) = 0 Rightarrow x=0, ; e^x = 0, ; or ; 3+x = 0. endalign* but $e^x eq 0$ for any type of $x in chathamtownfc.netbbR.$ Consequently, the second equation yields no solution. Therefore, our just solutions space $x=0$ and $x=-3.$
3. $ 5^2x=3 $
Solution. Here, us cannot usage the strategy used above, since there is no typical base in between $5$ and $3.$ So, we will certainly now check out the inverses the exponential functions, i m sorry will current us through a strategy to fix such problems!

Part 2. Logarithmic functions (the inverses that exponential functions!)

permit $b$ it is in a positive number such that $b eq 1.$ The logarithmic function through base $b,$ denoted through $log_b$ is identified by:$$ log_b(x) = y Leftrightarrow b^y = x$$
In other words:$log_b x= $ "the exponent $(y)$ that us raise the base $b$ to in order to obtain $x$"

Exercise. uncover $log_2 8$

eginalign*log_2 8 = k & Leftrightarrow 2^k = 8 \& Leftrightarrow k = log_2 8 = 3.endalign*
$f(x) =log_b x$ and also $g(x)= b^x$ space inverses! notification that in order to be inverses, the logarithmic and exponential attributes must have actually the same base $b$!
We will verify that $f(x) = log_b x$ and also $g(x)=b^x$ room inverses by utilizing the Cancellation Property:eginalign*f(g(x)) &= log_b(g(x)) \&= log_b(b^x) \&= x \& quad since ; log_b(b^x) = k Leftrightarrow b^k = b^x Leftrightarrow k=x \& quad extrm (by the 1-1 property)endalign*Similarly,eginalign*g(f(x)) &= b^left( f(x) ight) \&= b^log_b x \&= x endalign*since $log_b(x) = $ "the exponent such that once $b$ is elevated to it, it return the value $x$" by definition. i.e. $log_b(x) = k Leftrightarrow b^log_b x = b^k =x.$Graphically, we can see the they room inverses because the features reflect around the line $y=x.$ think about $f(x)= log_2(x)$ (in blue) and $g(x)=2^x$ (in red), whose graphs are provided below:

Notice that the domain that $y=b^x$ is the whole real line, i beg your pardon is now the range of $y= log_b x.$ Similarly, the variety of $y =b^x$ is equal to the domain of $y = log_b x,$ i m sorry is the term $(0,infty).$ This reflects that if $y=0$ was a horizontal asymptote because that $g(x)=b^x, ; x=0$ is currently a vertical asymptote because that $f(x)=log_b x.$

Part 3. Two crucial Logarithms


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We commonly write $log x$ to average $log_10 x.$ (i.e. The convention is come not compose the base of ten)
2. $b=e$
We typically write $ln x$ to mean $log_e x.$ "$ln$" stands for "natural logarithm" (so try not come say it favor "lawn", though plenty of do)
To include confusion, chathamtownfc.netematicians at greater levels regularly only take into consideration the natural logarithm, and so create $log x$ to mean base $e$ rather of basic $10.$ hopefully the base will certainly be clean from context in this situations.

Part 3. nature of Logarithms

$log_b b^x = x $
ii. $ log_b 1 =0 $
iii. $log_b b = 1 $
Convince yourself the these room true by merely using the definition of logarithms!In summary, we have actually that:$$ underbracelog_br =s_ extlogarithmic form qquad qquad way qquad qquad underbraceb^s=r_ extexponential form$$Recall the basic properties of features of the type $y=b^x$, wherein $b$ is a consistent positive real number:
1. $ b^r cdot b^s = b^r+s $
2. $ b^r div b^s = b^r-s $
3. $(b^r)^s = b^rs $
As formerly discussed, convert $x$ and also $y$ gives the inverse duty $y = log_b x.$ listed below are more properties the this function. (It might be valuable for you come make note of just how these nature are related to those of exponential functions, offered above.)
1. $ log_b(rcdot s) = log_b r + log_b s $
2. $ log_b(fracrs) = log_b r - log_b s $
3. $log_b(r) = fraclog_s rlog_s bqquad $ ("change of base" formula)
Below is the proof of (1). The remainder will be left to you as an exercise!Proof of (1):
Call $x=log_b r,$ $y=log_b s,$ and $z = log_b(rcdot s).$ We require to show that $z=x+y.$ Well, by definition of logarithms, we have actually the following:eginalign*x &= log_b r Leftrightarrow b^x = r \y &= log_b s Leftrightarrow b^y = s \z &= log_b(rcdot s) Leftrightarrow b^z = rcdot s.endalign*Therefore, $b^z = rcdot s = b^x b^y = b^x+y ;$ (by building 1 of exponents above)$Rightarrow b^z = b^x+y $$Rightarrow z = x+y; $ (by $1-1$ property of exponential functions)$Rightarrow log_b(rcdot s) = log_b r + log_b s ;$ together desired.

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Mini-Lecture. Listed below is a mini lecture around exponentials and logarithms.

Example. solve the following equations for $x$:

1. $ 5^2x = 3 ;$ (from very first example, #3, revisited)
Solution: Our score is to solve for $x$, which is an exponent at the moment. So, in order to get our hands at the exponent, us simply use the inverse of the exponential role with a basic of $5$(i.e. $log_5 x$) come both sides. This will enable us to usage the residential property $log_b b^x = x.$$$log_5(5^2x) = log_5(3) $$This gives $2x = log_5(3)$Finally, we have: $x=fraclog_5(3)2$ (which cannot be simplified further without a calculator!)
2. $ 4 (1+10^5x) = 404 $
Solution. A common approach is to very first isolate the exponential function. eginalign* 4(1+10^5x) = 404 &Rightarrow 1+10^5x = frac4044 \ &Rightarrow 1+10^5x = 101 \ &Rightarrow 10^5x = 100 endalign* Now, we have the right to either express both sides of the equation utilizing the exact same base, or use the station of the exponential function to both sides. Let"s proceed using the very first option, due to the fact that the same base the $10$ seems relatively obvious. eginalign* 10^5x=10^2 &Rightarrow 5x = 2 ; ext through 1-1 property \ &Rightarrow x = frac25 endalign*
3. $ e^x - 12 cdot e^-x - 1 = 0 $
Solution: This is merely a quadratic in disguise, i m sorry is revealed once we multiply both political parties by $e^x$ (to cancel with $e^-x$). eginalign* e^x - 12 cdot e^-x - 1 = 0 Rightarrow e^x(e^x) - 12 cdot e^-x(e^x) - 1(e^x) = 0(e^x) \ &Rightarrow (e^x)^2 - 12cdot e^0 - e^x = 0 \ &Rightarrow (e^x)^2 - e^x - 12 = 0 endalign* here is our quadratic! If this is tho unclear, simply let $z = e^x$. So, our identical equation is offered by: $(z)^2 - z - 12 = 0.$ This can be factored together follows: $$z^2 - z - 12 = (z-4)(z+3) = 0,$$ which offers that $z=4$ or $z=-3.$ currently we express everything in regards to $x$ in order to resolve the desired equation. We have actually $$e^x = 4 ext or e^x=-3 ; ext i m sorry isn"t true for any x!$$ So, there is just one valid solution, which we discover by applying $ln()$ come both sides of $e^x = 4:$ $$ln(e^x) = ln(4) Rightarrow x = ln(4).$$
4. $ log_3(x+29) - 2 log_3 (x-1)=0 $
Solution: We first notice that the logarithms have actually the same base, so our nature of Logarithms room applicable. Utilizing the $3$rd residential or commercial property (which allows us to "bring index number up or down") yields the following equivalent equation: $$ log_3(x+29) - log_3((x-1)^2) = 0.$$ currently we deserve to use the $2$nd property, which gives: $$log_3left( fracx+29(x-1)^2 ight) = 0.$$ The definition of logarithms states: $$ log_3left( fracx+29(x-1)^2 ight) = 0 Leftrightarrow 3^0 = fracx+29(x-1)^2.$$ Consequently, we simply need to deal with the following quadratic: eginalign* 1cdot (x-1)^2 = x+29 &Rightarrow x^2 - 2x +1 = x+29 \ &Rightarrow x^2 - 3x -28 = 0\ &Rightarrow (x-7)(x+4) = 0 \ &Rightarrow x = 7 ext or x = -4. endalign*
Note: we must recall that logarithmic features have restrictions on their domains. Therefore, prior to declaring our final solutions, we must inspect that the $x$-values don"t hurt the domain names of the original logarithmic functions. Substituting earlier into the initial problem, us have:
because that $x= 7:$ we show: LHS (left hand side) = RHS (right hand side) that the original equation. $$ extLHS = log_3 (7+29) - 2 log_3 (7-1) = log_3 left( frac366^2 ight) = log_3 ( 1) = 0 = ext RHS.$$ The vital thing to note is that as soon as we an initial substituted $x=7,$ both logarithmic features were defined! because that $x=-4:$$$ extLHS = log_3 (-4 +29) - 2 log_3 (-4-1) = log_3 left( 25 ight) - 2 log_3 ( -5).$$But $log_3(-5)$ is undefined! Therefore, $x=-4$ is not a valid solution!So $x=7$ is the only solution.