How Do Logarithms Simplify Exponential Equations and Factor Polynomials?

In summary, the conversation discusses the proof of the equation log a^x = x log a and its application in factoring polynomials. It also mentions searching for solutions among the integer divisors of the free term and using them to factorize the polynomial. The conversation concludes with a proposed factorization of the polynomial.
  • #1
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My book have a really crappy proof of how

log a^x = x log a

can be true. Can someone help me?


Another question deals with an application of it which made me really confused:

Factorise (4a^3) - (29 a^2) + 47a - 10

then solve (4*4^3x) - (29*4^2x) + 47*4^x - 10 = 0


I see many similarities between the two equations, but I don't know how to factorise polynomials. Plz help me! o:)
 
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  • #2
Link said:
My book have a really crappy proof of how

log a^x = x log a

can be true. Can someone help me?

What other elegant proof can be given,except the one which uses the definiton??

By "log" let's see a logaritm in an arbitrary base call it {b}.Since logarithm in the base "b" and the exponential "b" to the power of "x" are inverse functions,it follows simply that:

[tex] \log_{b} a^x= \log_{b} b^{(\log_{b} a)x}=x \log_{b} a [/tex]


Link said:
Another question deals with an application of it which made me really confused:
Factorise (4a^3) - (29 a^2) + 47a - 10
then solve (4*4^3x) - (29*4^2x) + 47*4^x - 10 = 0
I see many similarities between the two equations, but I don't know how to factorise polynomials. Plz help me! o:)


It's not always possible to factorize polynomials,since that would require finding all solutions.And for polynomials of degree larger than 4,that is generally impossible.For your case,since it has integer cofficients,try to searc solutions among the integer divisors of the free term,which is "-10".
You'll "2" as a root for the polynomial.From there,just bu dividing the original polynomial through "a-2",u'll left with a second degree polynomial whose roots are easy to find.
From there u can factorize all the polynomial:To chech your result,i say that it looks like that:
[tex] P(a)=(a-2)(a-5)(a-\frac{1}{4}) [/tex]

Daniel.
 
  • #3



Sure, I can help you with understanding logarithms and an application of it as well as factoring polynomials. Let's start with the concept of logarithms.

A logarithm is the inverse function of exponentiation. In simpler terms, it is a way to find the exponent that was used to get a certain number. For example, in the equation log base 2 of 8 = 3, the logarithm is asking "What power do I need to raise 2 to in order to get 8?" And the answer is 3.

Now, let's look at the equation log a^x = x log a. This equation is known as the power rule of logarithms. It states that the logarithm of a power is equal to the exponent multiplied by the logarithm of the base. This can be proven using the definition of logarithms and exponentiation. If you need further clarification on the proof, I suggest looking for online resources or asking your teacher for help.

Moving on to the application of logarithms, they are commonly used in various fields such as science, finance, and engineering. One example is in the field of earthquake measurement, where the Richter scale uses logarithms to measure the intensity of earthquakes. Another example is in finance, where logarithms are used to calculate compound interest and growth rates.

Now, let's talk about factoring polynomials. The first step in factoring is to look for common factors among the terms. In the given equation, we can see that all the terms have a common factor of a, so we can factor that out to get a(4a^2 - 29a + 47) - 10. Next, we need to factor the remaining polynomial, which can be done by finding two numbers that multiply to give 47 and add to give -29. These numbers are -1 and -47, so we can factor the polynomial as a(4a^2 - 47a - a + 47) - 10. Then, we can group the terms to get a((4a^2 - 47a) - (a - 47)) - 10. Finally, we can factor out the common terms in each group to get a(4a(a - 47) - 1(a - 47)) - 10. This can be further simplified to a(a - 47)(4a - 1) - 10.

To solve
 

FAQ: How Do Logarithms Simplify Exponential Equations and Factor Polynomials?

What is a logarithm?

A logarithm is the inverse function of exponentiation. It is used to determine the power to which a base number must be raised to produce a given number.

How is a logarithm written?

A logarithm is written as logb(x), where b is the base and x is the number for which the logarithm is being calculated.

What are some common applications of logarithms?

Logarithms are used in various fields such as mathematics, physics, chemistry, and engineering. Some common applications include calculating earthquake magnitude, measuring sound intensity, and determining pH level.

What is the difference between a natural logarithm and a common logarithm?

A natural logarithm (ln) uses a base of e (approximately 2.71828), while a common logarithm (log) uses a base of 10.

How can logarithms be used to solve exponential equations?

Logarithms can be used to solve exponential equations by taking the logarithm of both sides of the equation and using logarithm rules to simplify. This allows us to solve for the unknown variable in the exponent.

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