1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
puteri [66]
3 years ago
6

Express 3^4 = x as a logarithmic equation.

Mathematics
2 answers:
Bingel [31]3 years ago
5 0

Answer:  The required logarithmic equation is 4=\log_3x.

Step-by-step explanation:  We are given to write the following exponential equation as a logarithmic equation :

3^4=x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

We will be using the following property of logarithms :

a^b=c~~~~~~~~~~\Rightarrow b=\log_ac.

Therefore, from equation (i), we get

3^4=x\\\\\Rightarrow 4=\log_3x.

Thus, the required logarithmic equation is 4=\log_3x.

meriva3 years ago
3 0

Note that the base in both the exponential form of the equation and the logarithmic form of the equation (above) is "b<span>", but that the </span>x<span> and </span>y<span> 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.</span>

<span><span>Solve </span><span>log2(x) = 4</span>.</span>

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

<span><span><span>Solve </span><span>log2(</span></span>8<span><span>) = x</span>.</span></span>

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

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

<span>log2(8) = ln(8) / ln(2)</span>

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.

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

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":

<span>log2(x) + log2(x – 2) = 3</span> <span>
log2(4) + log2(4 – 2) ?=? 3</span> <span>
log2(4) + log2(2) ?=? 3</span>

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

<span>log2(4) + log2(2) ?=? 3</span> <span>
log2(2</span>2<span>) + log2(2</span>1) ?=? 3 <span>
2 + 1 ?=? 3</span> <span>
3 = 3</span>

The solution checks. Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

<span><span>Solve </span><span>log2(log2(x))   = 1.</span></span><span>This may look overly-complicated, but it's just another log equation. To solve this, I'll need to apply The Relationship twice:<span>log2(log2(x)) = 1</span> 
21 = <span>log2(x)</span> <span>
2 = log2(x)</span> <span>
x = 22</span> <span>
x = 4</span><span>Then the solution is </span><span>x = 4</span>.</span><span><span>Solve </span><span>log2(x2)  = (log2(x))2</span>.</span><span>First, I'll write out the square on the right-hand side:<span>log2(x2) = (log2(x))2</span> <span>
log2(x2) = (log2(x)) (log2(x))</span>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:<span>2log2(x) = [log2(x)] [log2(x)]</span> <span>
0 = [log2(x)] [log2(x)]  –  2log2(x)</span>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:<span>0 = </span><span>[log2(x)] [log2(x) – 2]</span> <span>
log2(x) = 0  or  log2(x) – 2 = 0</span> <span>
20 = x   or  log2(x) = 2</span> <span>
1 = x  or  22 = x</span> <span>
1 = x  or  4 = x</span><span><span>The solution is </span><span>x = 1, 4</span><span>.</span></span></span>
You might be interested in
16 is 64% of what number
Tems11 [23]

Answer:

25

Step-by-step explanation:

Is means equals and of means multiply

16 = 64% * W

Change to decimal form

16 = .64 * W

Divide each side by .64

16/.64 = .64W/.64

25 = W

7 0
3 years ago
Read 2 more answers
Solve the system using the substitution method. y = -5x – 13 6x + 6y = -6 please help me NO LINKS!
lana [24]

Answer:

Step-by-step explanation:

y=-5x-13

Since we know the value of y we can substitute it in

6x+6(-5x-13)=-6

6x-30x-78=-6

-24x=72

-x=3

x=-3

Now that we know the value of x we can solve Y

y=-5(-3)-13

y=15-13

y=2

3 0
3 years ago
Analyze the diagram, then solve for x.
Daniel [21]
9 because if you add 6x and 4x it makes 10x. if youre finding a 90 degree angle then it has to equal 90. 10 x 9 = 90.
7 0
3 years ago
Read 2 more answers
Answers<br> A. 5<br> B. 9<br> C. 11<br> D. 13
MA_775_DIABLO [31]
4k-7=2k+3
-2k. -2k

2k-7=3
+7. +7
2k=10
2÷2 2÷10
k=5

4(5)-7=13

Answer D
5 0
4 years ago
The width of a rectangular pool is 5 meters longer than the length. The perimeter of the pool is 58 meters. Use and equation to
garri49 [273]
The perimeter of a rectangular pool = 2l + 2w

If the width is 5m longer than the length, 

Then Perimeter = 2(l + 5) + 2l
                          = 4l + 10  {THE EQUATION}

Since Perimeter = 58 m
Then 4l + 10 = 58 m

⇒  l  =  12 m

Thus the Length of the Pool = 12 m
8 0
4 years ago
Other questions:
  • The function g is given by g(x) = -2(x+2)-6. Which of the following expresses the value of g(b-5)?
    13·1 answer
  • PLEASE HELP !!?!??!?!?!?!!
    12·1 answer
  • Maths Watch
    12·1 answer
  • Please whoever knows they can solve this, help me. I will be marking the branliest.
    8·2 answers
  • Kwame rode bicycle for a distance of xkm and walked for another 1/2 hour at a rate of 6km/hour. If Kwame covered a total distanc
    13·1 answer
  • Solve for x . Please help !
    5·1 answer
  • Need help ASAP!<br> Which inequality is represented by this graph
    12·1 answer
  • 10^-3<br> it told me i needed to write more characters
    11·2 answers
  • 3x+2x=20. What is the answer dor this equation. ​
    6·1 answer
  • Pleaseeee help me ………
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!