Example 5500 = 25500 * (0.85)^x

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.

interesting sites.

 

5500 divid. 25500 = 0.21

 

so log 0.21 divided by log 0.85 = x

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!

[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...

... 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.

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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.

no problem thanks

log with no base implies log₁₀

ln (natural log) is log_e