## Exponential and also Logarithmic Functions

**Part 1. Exponential functions **

The *exponential function*v base $b$ is identified for $x in jajalger2018.orgbbR$ by $f(x) = b^x,$ whereby $b > 0$ and also $b eq 1.$

**Graphically, we deserve to see the adhering to properties:**

￼￼￼￼￼Furthermore, due to the fact that $b > 0$ (by definition), we recognize that $f(x) = b^x eq 0$ for any $x in jajalger2018.orgbbR.$ Consequently, we have actually a

￼￼￼￼￼Furthermore, due to the fact that $b > 0$ (by definition), we recognize that $f(x) = b^x eq 0$ for any $x in jajalger2018.orgbbR.$ Consequently, we have actually a

*horizontal asymptote*in ~ the heat $y=0.$**The above properties of increasing and also decreasing show that exponential functions are $1-1,$ and therefore have actually inverses (which will certainly be debated in component 2).￼￼￼￼The**

allow $b$ be a hopeful number such that $b
eq 1.$ The

*natural exponential function*is known as $y=e^x,$ where $e$ is Euler’sirrational number: $2.71828cdots$

### Example. resolve the complying with exponential equations:

1. | $2^x4^3x+1 = frac2sqrt8 $ |

Solution: One strategy is to express both political parties in terms of the exact same base, specific $b=2,$ so the the nature of exponents have the right to 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 that $f(7x+2) = fleft( -frac12
ight),$ whereby $f(x) = 2^x,$ and because exponential functions are $1-1,$ we can conclude the $7x+2 = -frac12.$Therefore, $x = -frac514.$ | |

2. | $ 3x(e^x) + x^2 (e^x) = 0 $ |

Solution. Very first notice that $xcdot e^x$ is a common factor. eginalign* (x)(e^x)(3+x) = 0 Rightarrow x=0, ; e^x = 0, ; or ; 3+x = 0. endalign* yet $e^x eq 0$ for any kind of $x in jajalger2018.orgbbR.$ Consequently, the 2nd equation yields no solution. Therefore, our just solutions space $x=0$ and $x=-3.$ | |

3. | $ 5^2x=3 $ |

Solution. Here, we cannot usage the strategy used above, due to the fact that there is no typical base in between $5$ and $3.$ So, we will now check out the inverses the exponential functions, i m sorry will current us through a strategy to settle such problems! |

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

allow $b$ be a hopeful number such that $b
eq 1.$ The *inverses*that exponential functions!)

*logarithmic function*through base $b,$ denoted by $log_b$ is defined by:$$ log_b(x) = y Leftrightarrow b^y = x$$**In other words:$log_b x= $ "***the exponent**$(y)$ that we raise the basic $b$ to in stimulate to obtain $x$*"**$f(x) =log_b x$ and $g(x)= b^x$ space**

Note:### Exercise. find $log_2 8$

eginalign*log_2 8 = k & Leftrightarrow 2^k = 8 \& Leftrightarrow k = log_2 8 = 3.endalign*Note:

*inverses*! an alert that in order to be inverses, the logarithmic and exponential features must have the**same base**$b$!**We will certainly verify that $f(x) = log_b x$ and also $g(x)=b^x$ are inverses by making use of the Cancellation Property:eginalign*f(g(x)) &= log_b(g(x)) \&= log_b(b^x) \&= x \& quad because ; 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**together that once $b$ is increased 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 deserve to see the they room inverses because the functions reflect about the heat $y=x.$ consider $f(x)= log_2(x)$ (in blue) and $g(x)=2^x$ (in red), whose graphs are offered below:**

Notice the the domain of $y=b^x$ 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 interval $(0,infty).$ This mirrors that when $y=0$ was a horizontal asymptote because that $g(x)=b^x, ; x=0$ is currently a

To add confusion, jajalger2018.orgematicians at higher levels often

i.

Convince yourself the these space true by merely using the meaning of logarithms!Notice the the domain of $y=b^x$ 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 interval $(0,infty).$ This mirrors that when $y=0$ was a horizontal asymptote because that $g(x)=b^x, ; x=0$ is currently a

*vertical asymptote*for $f(x)=log_b x.$**Part 3. Two essential Logarithms **

1. You are watching: How to find inverse of exponential function | $b=10$ |

We commonly write $log x$ to mean $log_10 x.$ (i.e. The convention is to not write the base of ten) | |

2. | $b=e$ |

We normally write $ln x$ to mean $log_e x.$ "$ln$" means "natural logarithm" (so shot not to say it prefer "lawn", though numerous do) |

*only*consider the herbal logarithm, and also so create $log x$ to mean base $e$ rather of base $10.$ hope the base will be clean from context in these situations.**Part 3. properties of Logarithms **

i.$log_b b^x = x $ | |

ii. | $ log_b 1 =0 $ |

iii. | $log_b b = 1 $ |

**In summary, we have actually that:$$ underbracelog_br =s_ extlogarithmic form qquad qquad method qquad qquad underbraceb^s=r_ extexponential form$$Recall the basic properties of functions of the kind $y=b^x$, wherein $b$ is a consistent positive actual number:**

As previously discussed, convert $x$ and also $y$ offers the inverse function $y = log_b x.$ listed below are more properties of this function. (It might be beneficial for you come make keep in mind of exactly how these nature are related to those the exponential functions, given above.)

Below is the proof of (1). The rest will be left come you together an exercise!Proof the (1): 1. | $ b^r cdot b^s = b^r+s $ |

2. | $ b^r div b^s = b^r-s $ |

3. | $(b^r)^s = b^rs $ |

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) |

**Call $x=log_b r,$ $y=log_b s,$ and $z = log_b(rcdot s).$ We require to display 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 index number above)$Rightarrow b^z = b^x+y $$Rightarrow z = x+y; $ (by $1-1$ building of exponential functions)$Rightarrow log_b(rcdot s) = log_b r + log_b s ;$ together desired.**

See more: Reduce 18/24 In Simplest Form Of 18 24? Simplify Or Reduce Fraction 18/24

See more: Reduce 18/24 In Simplest Form Of 18 24? Simplify Or Reduce Fraction 18/24

**Mini-Lecture. Listed below is a mini lecture around exponentials and logarithms.**### Example. settle the following equations because that $x$:

1. | $ 5^2x = 3 ;$ (from first example, #3, revisited) |

Solution: Our score is to solve for $x$, which is one exponent in ~ the moment. So, in bespeak to obtain our hands at the exponent, us simply use the train station of the exponential duty with a base of $5$(i.e. $log_5 x$) come both sides. This will enable us to usage the building $log_b b^x = x.$$$log_5(5^2x) = log_5(3) $$This gives $2x = log_5(3)$Finally, us have: $x=fraclog_5(3)2$ (which can not be simplified further without a calculator!) | |

2. | $ 4 (1+10^5x) = 404 $ |

Solution. A common strategy 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 refer both political parties of the equation making use of the exact same base, or apply the train station of the exponential function to both sides. Let"s proceed using the an initial option, since the very same base of $10$ seems relatively obvious. eginalign* 10^5x=10^2 &Rightarrow 5x = 2 ; ext by 1-1 property \ &Rightarrow x = frac25 endalign* | |

3. | $ e^x - 12 cdot e^-x - 1 = 0 $ |

Solution: This is simply a quadratic in disguise, which is revealed when 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* below is our quadratic! If this is still unclear, simply let $z = e^x$. So, our indistinguishable equation is provided by: $(z)^2 - z - 12 = 0.$ This can be factored as follows: $$z^2 - z - 12 = (z-4)(z+3) = 0,$$ which provides that $z=4$ or $z=-3.$ currently we express every little thing in regards to $x$ in bespeak to deal with the wanted 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 only one precious solution, which we discover by using $ln()$ to 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 very first notice the the logarithms have the very same base, for this reason our properties of Logarithms space applicable. Using the $3$rd residential property (which permits us come "bring index number up or down") yields the following equivalent equation: $$ log_3(x+29) - log_3((x-1)^2) = 0.$$ now we deserve to use the $2$nd property, i beg your pardon gives: $$log_3left( fracx+29(x-1)^2
ight) = 0.$$ The an interpretation of logarithms states: $$ log_3left( fracx+29(x-1)^2
ight) = 0 Leftrightarrow 3^0 = fracx+29(x-1)^2.$$ Consequently, we merely need to solve the adhering to 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 have to recall that logarithmic features have limitations on your domains. Therefore, before declaring our last solutions, we must inspect that the $x$-values don"t hurt the domain names of the initial logarithmic functions. Substituting earlier into the initial problem, we have: for $x= 7:$ we show: LHS (left hand side) = RHS (right hand side) of 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 necessary thing to note is that once we an initial substituted $x=7,$ both logarithmic functions 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. |