Simple algebraic / logarithmic question

In summary, the equation 1.048576 = (1+m)^12 can be simplified to 1.024^2 = (1+m)^{12}. Taking the 12th root of 1.048576 results in 1.003961.
  • #1
mofoj
5
0

Homework Statement



(1.024)^2 = (1 + m)^12

Homework Equations





The Attempt at a Solution



1.048576 = (1+m)^12
I tried to do a square root of 12 for both sides but does not appear to be correct. I've also tried doing log to bring the exponents to the front for both sides, but something is wrong. I learned this long time ago so I really can't remember how to approach it.
 
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  • #2
I guess you are trying to solve for m?

Starting with a 12th root of both sides seems like a reasonable way of dealing with the exponent.
 
  • #3
Borek said:
I guess you are trying to solve for m?

Starting with a 12th root of both sides seems like a reasonable way of dealing with the exponent.


Yeah that's what I did, but I did not get the correct answer, which should be 0.003961.
 
  • #4
What did you get when you took the 12th root of 1.048576? And did you subtract 1 from both sides after taking the 12th root?
 
  • #5
eumyang said:
What did you get when you took the 12th root of 1.048576? And did you subtract 1 from both sides after taking the 12th root?

yes I did subtract the 1 after, but it is not the correct answer still.
 
  • #7
Mark44 said:
Show us what you did...

1.048576 = (1+m)^12

12th sq root
10.69516372 = 1 + m
9.69516372 = m


supposed to be 0.003961
 
  • #8
mofoj said:
1.048576 = (1+m)^12

12th sq root
10.69516372 = 1 + m
9.69516372 = m


supposed to be 0.003961

There's no such thing as a "12th sq root". You raised the left side to the 12th power. That's different from taking the 12th root, which is the same as the 1/12 th power.

1.048576 = (1+m)^12
(1.048576)^(1/12) = 1+m
 
  • #9
Mark44 said:
There's no such thing as a "12th sq root". You raised the left side to the 12th power. That's different from taking the 12th root, which is the same as the 1/12 th power.

1.048576 = (1+m)^12
(1.048576)^(1/12) = 1+m


finally! great, thanks for the help.
 
  • #10
Mark44 said:
There's no such thing as a "12th sq root".
A very good point. Unfortunately, even the "LaTex" used on this board requires that we enter a "12th root" as \sqrt[12]{x} !

You raised the left side to the 12th power. That's different from taking the 12th root, which is the same as the 1/12 th power.

1.048576 = (1+m)^12
(1.048576)^(1/12) = 1+m
 
  • #11
You can simplify things a bit by doing this:

[tex]1.024^2 = (1 + m)^{12}[/tex]

[tex]\left( 1.024^2 \right) ^{1/12} = \left[ (1 + m)^{12} \right]^{1/12}[/tex]

[tex]1.024^{1/6} = 1 + m[/tex]

[tex]\sqrt[6]{1.024} = 1 + m[/tex]

[tex]m = \sqrt[6]{1.024} - 1[/tex]
At least it keeps the numbers nice and tidy until the end!
 

1. What is algebra and why is it important?

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating those symbols to solve equations and represent relationships between quantities. It is important because it provides a foundation for higher level math and is used in various fields such as science, engineering, finance, and economics.

2. What is a variable and how is it used in algebra?

A variable is a symbol that represents an unknown value or quantity in an equation. It is used in algebra to help us solve problems and find solutions to equations by substituting different values for the variable and solving for the unknown value.

3. What is the difference between a linear and logarithmic equation?

A linear equation has a constant rate of change and can be represented by a straight line on a graph. A logarithmic equation, on the other hand, has a variable rate of change and can be represented by a curved line on a graph. In other words, a linear equation has a constant slope while a logarithmic equation has a changing slope.

4. How do you solve a simple algebraic equation?

To solve a simple algebraic equation, you need to isolate the variable on one side of the equation by using inverse operations. This means whatever operation is being done to the variable, you do the opposite operation to both sides of the equation. Once the variable is isolated, you can solve for its value by performing any necessary calculations.

5. What are the properties of logarithms?

The properties of logarithms include the product rule, quotient rule, power rule, and change of base rule. The product rule states that the logarithm of a product is equal to the sum of the logarithms of each factor. The quotient rule states that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and denominator. The power rule states that the logarithm of a power is equal to the exponent multiplied by the logarithm of the base. The change of base rule states that any logarithm can be rewritten in terms of a different base using the formula logb(x) = loga(x)/loga(b).

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