Logs question (1 Viewer)

[mtsmahia]

Sorry for another question on same topic..

Solve x using logarithms

1)
2^x = 5^x-1

 

[lyounamu]

Sorry for another question on same topic..

Solve x using logarithms

1)
2^x = 5^x-1

2^x = 5^(x -1)
x = ln2(5^(x-1))
x = (x-1)(ln2(5))
x/(x-1) = ln2(5)
x/(x-1) = 2.32..
x = 2.32.... x - 2.32....
2.32... = 1.32... x
x = 1.756...

 

[lyounamu]

log both sides and solve for x

Duh. Everyone knows that. It is HOW they come about solving it.

How helpful.

 

[mtsmahia]

log both sides and solve for x

yeah i know---but how

i dont know what to do after--

x log(10)2 = (x-1) log(10)5

 

[lyounamu]

if you know the method, they shouldn't have big problem except silly mistakes

Do you know what? Most Maths questions are like that. You can at least start with some formula and stuff but it is HOW part where they get stuck.

And for some people, even small problem can stumble them over.

 

[mtsmahia]

2^x = 5^(x -1)
x = ln2(5^(x-1))
x = (x-1)(ln2(5))
x/(x-1) = ln2(5)
x/(x-1) = 2.32..
x = 2.32.... x - 2.32....
2.32... = 1.32... x
x = 1.756...

whats In--we have learnt it

 

[lyounamu]

whats In--we have learnt it

ln = log base e.

I am sure you have learnt this.

Because you can change this: log2(5) to loge(5)/loge(2) (or ln5/ln2) by the change of base law.