The exponential function
$f\left(x\right)={b}^{x}$ is one-to-one, with domain
$(\text{\u2212}\infty ,\infty )$ and range
$\left(0,\infty \right).$ Therefore, it has an inverse function, called the
logarithmic function with base$b.$ For any
$b>0,b\ne 1,$ the logarithmic function with base
b , denoted
${\text{log}}_{b},$ has domain
$(0,\infty )$ and range
$\left(\text{\u2212}\infty ,\infty \right),$ and satisfies
${\text{log}}_{b}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}{b}^{y}=x.$
The most commonly used logarithmic function is the function
${\text{log}}_{e}.$ Since this function uses natural
$e$ as its base, it is called the
natural logarithm . Here we use the notation
$\text{ln}(x)$ or
$\text{ln}\phantom{\rule{0.1em}{0ex}}x$ to mean
${\text{log}}_{e}\left(x\right).$ For example,
and their graphs are symmetric about the line
$y=x$ (
[link] ).
At this
site you can see an example of a base-10 logarithmic scale.
In general, for any base
$b>0,b\ne 1,$ the function
$g\left(x\right)={\text{log}}_{b}(x)$ is symmetric about the line
$y=x$ with the function
$f\left(x\right)={b}^{x}.$ Using this fact and the graphs of the exponential functions, we graph functions
${\text{log}}_{b}$ for several values of
$b>1$ (
[link] ).
Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of logarithms.
Rule: properties of logarithms
If
$a,b,c>0,b\ne 1,$ and
$r$ is any real number, then
Now we can solve the quadratic equation. Factoring this equation, we obtain
$\left({e}^{x}-3\right)\left({e}^{x}-2\right)=0.$
Therefore, the solutions satisfy
${e}^{x}=3$ and
${e}^{x}=2.$ Taking the natural logarithm of both sides gives us the solutions
$x=\text{ln}\phantom{\rule{0.1em}{0ex}}3,\text{ln}\phantom{\rule{0.1em}{0ex}}2.$
The solution is
$x={10}^{4\text{/}3}=10\sqrt[3]{10}.$
Using the power property of logarithmic functions, we can rewrite the equation as
$\text{ln}\left(2x\right)-\text{ln}\left({x}^{6}\right)=0.$ Using the quotient property, this becomes
$\text{ln}\left(\frac{2}{{x}^{5}}\right)=0.$
Therefore,
$2\text{/}{x}^{5}=1,$ which implies
$x=\sqrt[5]{2}.$ We should then check for any extraneous solutions.
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base.
a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec
Pls help me solve
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation
your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
Adri
am sorry for disturbing I really want to know math that's why
*I want to know the meaning of those symbols in sets*
e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis
*in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions
what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
Adri
I dont understand what you wanna say by (A' n B^c)^c'
Adri
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Adri
Ok so the set is formed by vectors and not numbers
Adri
A vector of length n
Adri
But you can make a set out of matrixes as well
Adri
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
Wait what's your math level?
Adri
High-school?
Adri
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
Adri
pls answer this question
*in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
Adri
I would say 24
Adri
Offer both
Adri
Sorry 20
Adri
Actually you have 40 - 4 =36 who offer maths or physics or both.
Adri
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
Adri
56-36=20 who give both courses... I would say that
Adri
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have:
n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation
40 = 24 + 32 - n(M and P) + 4
Now solve for the unknown using algebra:
40 = 24 + 32+ 4 - n(M and P)
40 = 60 - n(M and P)
Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P)
Solution:
n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form:
Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
the domain of the function is all real number excluding zero, because the rational function 1/x is a representation of a fractional equation (precisely inverse function). As in elementary mathematics the concept of dividing by zero is nonexistence, so zero will not make the fractional statement
Mckenzie
a function's answer/range should not be in the form of 1/0 and there should be no imaginary no. say square root of any negative no.
(-1)^1/2
Roha
domain means everywhere along the x axis. since this function is not discontinuous anywhere along the x axis, then the domain is said to be all values of x.
Andrew
Derivative of a function
Waqar
right andrew ... this function is only discontinuous at 0
Roha
of sorry, I didn't realize he was taking about the function 1/x ...I thought he was referring to the function x^3-2x.
Andrew
yep...it's 1/x...!!!
Roha
true and cannot be apart of the domain that makes up the relation of the graph y = 1/x. The value of the denominator of the rational function can never be zero, because the result of the output value (range value of the graph when x =0) is undefined.
Mckenzie
👍
Roha
Therefore, when x = 0 the image of the rational function does not exist at this domain value, but exist at all other x values (domain) that makes the equation functional, and the graph drawable.