Help please thanks very much

Express 3^4 = x as a logarithmic equation.

meriva

Note that the base in both the exponential form of the equation and the logarithmic form of the equation (above) is "b", but that the x and y switch sides when you switch between the two equations. If you can remember this — that whatever had been the argument of the log becomes the "equals" and whateverhad been the "equals" becomes the exponent in the exponential, and vice versa — then you should not have too much trouble with solving log equations.

Solve log2(x) = 4.

Since this is "log equals a number", rather than "log equals log", I can solve by using The Relationship:
 log2(x) = 4 
24 = x
16 = x

Solve log2(8) = x.

I can solve this by converting the logarithmic statement into its equivalent exponential form, using The Relationship:log2(8) = x
2 x = 8But 8 = 23, so:2 x = 23
x = 3

Note that this could also have been solved by working directly from the definition of a logarithm: What power, when put on "2", would give you an 8? The power 3, of course!

If you wanted to give yourself a lot of work, you could also do this one in your calculator, using the change-of-base formula:

log2(8) = ln(8) / ln(2)

Plug this into your calculator, and you'll get "3" as your answer. While this change-of-base technique is not particularly useful in this case, you can see that it does work. (Try it on your calculator, if you haven't already, so you're sure you know which keys to punch, and in which order.) You will need this technique in later problems.

Solve log2(x) + log2(x – 2) = 3I can't do anything yet, because I don't yet have "log equals a number". So I'll need to use log rules to combine the two terms on the left-hand side of the equation:log2(x) + log2(x – 2) = 3 
log2((x)(x – 2)) = 3 
log2(x2 – 2x) = 3Then I'll use The Relationship to convert the log form to the corresponding exponential form, and then I'll solve the result:log2(x2 – 2x) = 3 
23 = x2 – 2x 
8 = x2 – 2x 
0 = x2 – 2x – 8 
0 = (x – 4)(x + 2) 
x = 4, –2But if x = –2, then "log2(x)", from the original logarithmic equation, will have a negative number for its argument (as will the term "log2(x – 2)"). Since logs cannot have zero or negative arguments, then the solution to the original equation cannot be x = –2.The solution is x = 4.

Keep in mind that you can check your answers to any "solving" exercise by plugging those answers back into the original equation and checking that the solution "works":

log2(x) + log2(x – 2) = 3 
log2(4) + log2(4 – 2) ?=? 3 
log2(4) + log2(2) ?=? 3

Since the power that turns "2" into "4" is 2 and the power that turns "2" into "2" is "1", then we have:

log2(4) + log2(2) ?=? 3 
log2(22) + log2(21) ?=? 3 
2 + 1 ?=? 3 
3 = 3

Solve log2(log2(x)) = 1.This may look overly-complicated, but it's just another log equation. To solve this, I'll need to apply The Relationship twice:log2(log2(x)) = 1
21 = log2(x) 
2 = log2(x) 
x = 22 
x = 4Then the solution is x = 4.Solve log2(x2) = (log2(x))2.First, I'll write out the square on the right-hand side:log2(x2) = (log2(x))2 
log2(x2) = (log2(x)) (log2(x))Then I'll apply the log rule to move the "squared", from inside the log on the left-hand side of the equation, out in front of that log as a multiplier. Then I'll move that term to the right-hand side:2log2(x) = [log2(x)] [log2(x)] 
0 = [log2(x)] [log2(x)] – 2log2(x)This may look bad, but it's nothing more than a factoring exercise at this point. So I'll factor, and then I'll solve the factors by using The Relationship:0 = [log2(x)] [log2(x) – 2] 
log2(x) = 0 or log2(x) – 2 = 0 
20 = x or log2(x) = 2 
1 = x or 22 = x 
1 = x or 4 = xThe solution is x = 1, 4.

3 0

3 years ago

Read 2 more answers

Worth lot of points :))))

maxonik [38]

Step-by-step explanation:

slope = (y2 - y1)/(x2 - x1)

-4/3 = (6 - y)/[4-(-2)]

-4/3 = (6-y)/6

(-4/3)×6 = 6-y

-8 = 6-y

y = 6 + 8

= 14

7 0

3 years ago

Read 2 more answers

Write an equation of the line passing through the point (1,5) that is parallel to the line y= -3x+5

Semenov [28]

Answer:

y=-3x+8

Step-by-step explanation:

y=mx+5, m=-3, x=1, y=5

5=(-3)(1)+b

5=-3+b

8=b

4 0

3 years ago

(1, -4), y = -3x - 3

Alenkasestr [34]

Answer:

Step-by-step explanation:

Hope the graph helps

Mark me as brainiest pls begging you

7 0

3 years ago

Let 5 be the region that lies between the curves y=− xm ; y= − xn ; 0 < x < 1 where m and n are integers with 0 < n , m

Lena [83]

Answer:

(a) Please see the first figure attached

(b) The coordinates of the centroid are $G(\frac{2}{3}, \frac{m+n}{3} )$

(c) due to the definition of the centroid of a triangle, this will always lie inside the triangle, therefore, for any value of $m$ and $n$ , the centroid will lie

Step-by-step explanation:

Hi, let us first solve part (a). Since for any given values of $n$ and $m$ we will obtain two linear functions:

$y=-mx$ and

$y=-nx$

with $0\leq x\leq 1$ we can assure that our region is going to be a triangle. To see this, please take a look at the plot I generated using Wolfram. In this case, I have used two specific values for m and n but keeping the condition $0\leq n\leq m$ .

Now, for part (b) let me start remembering what the centroid is: the centroid of a triangle is the point where the three medians of the triangle meet. And a median of a triangle is a line segment from one vertex to the midpoint on the opposite side of the triangle (see the second figure where the medians are depicted in red and the centroid of the triangle, G is depicted in blue). For a given triangle $\bigtriangleup \rm{ABC}$ , the coordinates of its centroid $G$ are given by:

$G_x=\frac{A_x+B_x+C_x}{3}$ and $G_y=\frac{A_y+B_y+C_y}{3}$

Now let's apply this to our problem. Take a look at the first figure. The vertex A has clearly coordinates $(0,0)$ for any value of $m$ and $n$ since the two lines have their intersection with y-axis in this point.

To obtain the coordinates of $B$ and $C$ , let's use the given functions and the fact that the coordinate x is limited to 1. Then, we have:

For A:

$y=-mx$ then, when $x=1$ , substituting in the formula $y=-m$

For B and doing the same as for A:

$y=-nx$ then, when $x=1$ , substituting in the formula $y=-n$

Thus, the coordinates of the vertices of the triangle are: $A(1, m)$ , $B(1,3)$ and $C(0,0)$ and the coordinates of the centroid are:

$G_x=\frac{A_x+B_x+C_x}{3} = G_x=\frac{1+1+0}{3}\\G_x=\frac{2}{3}$

and

$G_y=\frac{A_y+B_y+C_y}{3}=\frac{m+n+0}{3}\\G_y=\frac{m+n}{3}$ .

Summarizing: the coordinates of the centroid of the region are $G(\frac{2}{3}, \frac{m+n}{3} )$

Now, for part (c), due to the definition of the centroid of a triangle, this will always lie inside the triangle, therefore, for any value of $m$ and $n$ , the centroid will lie. Other important points of the triangle, like the orthocentre and circumcentre, can lie outside in obtuse triangles. In right triangles, the orthocentre always lies at the right-angled vertex.

8 0

3 years ago