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Basic Log Function, Solve for X

  • Thread starter matadorqk
  • Start date
  • #1
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**Im helping a friend go through this problem, so "the answer will be shown eventually" because im 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 lets start
APPROACH #1
Ok, lets 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]
 
Last edited:

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
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
492
1
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
96
0
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
cristo
Staff Emeritus
Science Advisor
8,107
73
Ok, lets 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.
 

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