Basic Log Function, Solve for X

In summary: Also, this is incorrect: "So, we pass log 2^{x-1} to the other side, by adding on each side." This is incorrect because logarithms do not work that way. You should say "So, we pass log 2^{x-1} to the other side, by adding on each side to get x log 2."
  • #1
matadorqk
96
0
**Im helping a friend go through this problem, so "the answer will be shown eventually" because I am editing as we go through.

Homework Statement


[tex]2^{x-1}=25[/tex]

Homework Equations



All Log/Exponent Formulas, they'll be shown as we go.

The Attempt at a Solution



There are various ways of approaching this problem, so let's start
APPROACH #1
Ok, let's multiply both sides by [tex]\log_{10}[/tex]

So, [tex]\log 2^{x-1}=log 25 [/tex]

We know that [tex] \log a^{b}=b \log a [/tex], so apply this to our left hand side.

So, if we know the above formula, the [tex]\log 2^{x-1}=log a ^{b}[/tex]. So, we find our a and b.
A=2 and B=x-1

So, [tex]log(2)^{x-1}=(x-1)log2[/tex]

Therefore, [tex] (x-1) log 2 = log 25 [/tex] so multiply x-1.

Therefore, [tex] x log 2 - 1 log 2 = log 25 [/tex]

So we solve for x!

So, we pass log 2 to the other side, by adding on each side.

[tex] x log 2 = log 25 + log 2. [/tex]

Divide both sides by log 2.

[tex] x= \frac{log 25 + log 2}{log 2} [/tex]

Now use the calculator.

x=5.64

APPROACH #2
If we know that [tex](a^{b})(a^{c}) = a^{b+c}[/tex]

we know that [tex]2^{x-1}=(2^{x})(2^{-1}).[/tex]

Therefore, [tex](2^{x})(\frac{1}{2})=25[/tex].

So multiply both sides by 2, to cancel out the 1/2.

[tex] 2^{x}=50 [/tex]

Now its simple log work, we multiply both sides by [tex]log_{10}[/tex]

So [tex] log 2^{x} = log 50 [/tex]

So [tex] x log 2 = log 50 [/tex]

So [tex] x=\frac{log 50}{log 2} = 5.64 [/tex]
 
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  • #2
well from [tex]lg2^{x-1}=lg25[/tex] applying your law you'd get [tex](x-1)lg2=lg25[/tex] right?

and well lg2 and lg25 are constants...so how do you think you'd get x from here?
 
  • #3
in (x-1)log2=log25 why don't you just divide by log2 and find x?

did the answer have to be in base of 10?

if it could be any log base it would have been easier to do log base 2 to start off, which would give you the exact same answer. changing log base 2 (25) to log base 10 you get what you got.Edit: those are correct, but just a suggestion when you had (x-1)log 2= log 25 you could just divide and move the 1 over to the RHS to get: x=(log25)/(log2)+1, which again is exactly what you got just a little thing that might save you some time in a test.
 
Last edited:
  • #4
bob1182006 said:
in (x-1)log2=log25 why don't you just divide by log2 and find x?

did the answer have to be in base of 10?

if it could be any log base it would have been easier to do log base 2 to start off, which would give you the exact same answer. changing log base 2 (25) to log base 10 you get what you got.


Edit: those are correct, but just a suggestion when you had (x-1)log 2= log 25 you could just divide and move the 1 over to the RHS to get: x=(log25)/(log2)+1, which again is exactly what you got just a little thing that might save you some time in a test.

Yeah, attempt 2 is how i would do it, but he hasn't learned that so i tried to do it the 'long way' so he would understand better, you guys are right though.
 
  • #5
matadorqk said:
Ok, let's multiply both sides by [tex]\log_{10}[/tex]
You've written this twice, however it is incorrect. It makes no more sense to say "multiply both sides by log" than it does to say "multiply both sides by e." log denotes the logarithm function, and so here you are taking the logarithm of both sides of the equation.
 

1. What is the basic log function?

The basic log function is a mathematical function that is the inverse of the exponential function. It is written as logb(x), where b is the base and x is the argument. This function tells us what power the base needs to be raised to in order to get the argument value.

2. How do I solve for x in a log function?

To solve for x in a log function, you need to isolate the variable by using the properties of logarithms. These include the power rule, product rule, quotient rule, and the change of base formula. Once you have isolated the variable, you can solve for it by taking the antilog of both sides.

3. Can a log function have a negative argument?

No, a log function cannot have a negative argument. This is because the logarithm of a negative number is undefined. The argument of a log function must always be a positive number to have a real solution.

4. What is the difference between natural log and common log?

The natural log, written as ln(x), has a base of e, which is an irrational number approximately equal to 2.71828. The common log, written as log(x), has a base of 10. This means that the natural log is the logarithm of a number to the base e, while the common log is the logarithm of a number to the base 10.

5. Can a log function have a base of 1?

No, a log function cannot have a base of 1. This is because the logarithm of 1 is always equal to 0, so the function would not have a unique solution. Additionally, any number raised to the power of 0 is equal to 1, so the equation would become x = 1, which is not a logarithmic equation.

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