How do you find x when it's an exponent

[CrazCo]

Example 5500 = 25500 * (0.85)^x

[mooeypoo]

Do you know about logarithms? I don't want to expose the answer straight-out, but you might want to read a bit about them

 

http://en.wikipedia.org/wiki/Logarithm and http://mathworld.wolfram.com/Logarithm.html are good sources to get you started.

[CrazCo]

interesting sites.

 

5500 divid. 25500 = 0.21

 

so log 0.21 divided by log 0.85 = x

[mooeypoo]

What is the base of your logs?

 

Remember,

 

[math]x = b^y[/math] means that [math]y = \log _b{x}[/math]

 

and check this out:

http://en.wikipedia.org/wiki/Logarithm#Change_of_base

 

but you're definitely on the right track!

[Cap'n Refsmmat]

[math]5500 = 25500 \cdot 0.85^x[/math]

 

[math]0.21 = 0.85^x[/math]

 

[math]\log_{0.85} 0.21 = x[/math]

 

[math]\frac{\log 0.21}{\log 0.85} = x[/math]

 

Looks like CrazCo already got that bit. Doesn't matter what the bases are, it's still the right answer...

[mooeypoo]

... when you write "log" with no base, it's as if the base is e, isn't it?

 

In his case the base isn't e, it's a number. It seemed to me to be important.. I might be going ahead of myself here, though, in which case, I apologize.

---

Merged post follows:

Consecutive posts merged

OH *SHOOT!* I missed the part where you translated the "base of" the log to log divided by log..

 

I actually missed my own advice! ha!

 

You're absolutely right, I apologize, the answer's correct as is.

 

Sorry, CrazCo.

[CrazCo]

no problem thanks

[Klaynos]

log with no base implies log₁₀

ln (natural log) is log_e