To check the result, substitute
$\text{\hspace{0.17em}}x=10\text{\hspace{0.17em}}$ into
$\text{\hspace{0.17em}}\mathrm{log}\left(3x-2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right).$
Using the one-to-one property of logarithms to solve logarithmic equations
For any algebraic expressions
$\text{\hspace{0.17em}}S\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}T\text{\hspace{0.17em}}$ and any positive real number
$\text{\hspace{0.17em}}b,$ where
$\text{\hspace{0.17em}}b\ne 1,$
Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.
Given an equation containing logarithms, solve it using the one-to-one property.
Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form
$\text{\hspace{0.17em}}{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T.$
Use the one-to-one property to set the arguments equal.
Solve the resulting equation,
$\text{\hspace{0.17em}}S=T,$ for the unknown.
Solving an equation using the one-to-one property of logarithms
Solving applied problems using exponential and logarithmic equations
In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.
One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its
half-life .
[link] lists the half-life for several of the more common radioactive substances.
Substance
Use
Half-life
gallium-67
nuclear medicine
80 hours
cobalt-60
manufacturing
5.3 years
technetium-99m
nuclear medicine
6 hours
americium-241
construction
432 years
carbon-14
archeological dating
5,715 years
uranium-235
atomic power
703,800,000 years
We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
Someone should please solve it for me
Add 2over ×+3 +y-4 over 5
simplify (×+a)with square root of two -×root 2 all over a
multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15
Second one, I got Root 2
Third one, I got 1/(y to the fourth power)
I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
graph the following linear equation using intercepts method.
2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b
you were already given the 'm' and 'b'.
so..
y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line.
where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2
2=3x
x=3/2
then .
y=3/2X-2
I think
Given
co ordinates for x
x=0,(-2,0)
x=1,(1,1)
x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
I've run into this:
x = r*cos(angle1 + angle2)
Which expands to:
x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2))
The r value confuses me here, because distributing it makes:
(r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1))
How does this make sense? Why does the r distribute once
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
Brad
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis
vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As
'f(x)=y'.
According to Google,
"The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
GREAT ANSWER THOUGH!!!
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks.
"Â" or 'Â' ... Â