Hi, I have a problem. Its logarithms.

[Tikoonmunci]

325*(0.8)^t=5

Can anyone help me please?

 

[Office_Shredder]

You should probably take a logarithm. What are you having trouble with? You have to explain where you're getting stuck

 

[HallsofIvy]

325*(0.8)^t=5

Can anyone help me please?

Since you say "it's logarithms", you probably already know what you should do- the whole point of logarithms is that they "inverse" to exponentials: log(a^x)= x log(a), getting the "x" out of the exponent.

Rather that take the logarithm immediately (which is possible), I would reduce this to a "bare" exponential by dividing both sides by 325: (0.8)^t= 5/325= 1/65. Now take the logarithm of both sides. (Any logarithm will do. Common log (base 10) or natural log (base e), which are on a calculator, will give the same answer.)

 

[Susanne217]

325*(0.8)^t=5

Can anyone help me please?

Like in any equation you have isolate the unknown.

This looks to be High School level. You are properly familar with the function

$$y(t) = b \cdot a^t$$ ?

Then if you have to find t you use standard elimation such as : $$\frac{y(t)}{b} = a^t$$

take the logaritm on both sides of equality and use this fact to isolate t (remember $$a^t = t \cdot log(a))$$

Then the eqn is solved my friend :)

 

[Mentallic]

$$a^t = t \cdot log(a))$$

Correction, $$log(a^t)=tlog(a)$$

 

[Susanne217]

Correction, $$log(a^t)=tlog(a)$$

bang you got me chief :D

 

[Mentallic]

I'll let you off this time, but you just better be getting the heck on outta here and never let me see your face around these parts of town again, you hear!?

 

[Susanne217]

I'll let you off this time, but you just better be getting the heck on outta here and never let me see your face around these parts of town again, you hear!?

Well I didn't know there was a new sherif in these here parts of the Intermerwebs :)

 

[Mentallic]

Actually I'm still a chief. I keep civility by day, and smoke my peace pipe by night. Oh and I happen to help with homework every now and again