4000(1+(7.5/100))^x=4622,5: How do I solve this ?

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Zulgok

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Hello, I need to find x here, been trying for a while, no good. Any help appreciated.

4000(1+(7.5/100))^x=4622,5.
 
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We havent ever studied anything about logarithms, so no idea how to use it.
 
Zulgok said:
We havent ever studied anything about logarithms, so no idea how to use it.
Click to expand...

Then what have you been taught about exponential functions - and the inverse of such functions.

Otherwise you cannot solve these types of equations - may be use trial and error, but that is a long shot.
 
But this doesnt need any functions, other classmates done it without it. About x, the only thing i know is that it has to be 2.
 
Zulgok said:
Hello, I need to find x here, been trying for a while, no good. Any help appreciated.
4000(1+(7.5/100))^x=4622,5.
Click to expand...

The question looks like a 'How many years does it take $4000 to grow to $4622.50 at a compounded 7.5% interest per year'. So, if you don't know logarithms, just start working, or as Subhotosh Khan said, use trial and error:
at the end of year 1: 4000*1.075=4300
at the end of year 2: 4300*1.075=...
 
Zulgok said:
I need to find x here:

. . . . .\(\displaystyle 4000\, \left(\, 1\, +\, \dfrac{7.5}{100}\,\right)^x\, =\, 4622.5\)

We havent ever studied anything about logarithms....

....this doesnt need any functions, other classmates done it without it. About x, the only thing i know is that it has to be 2.

I used [trial and error] in test, but I didnt get all points for it.
Click to expand...
Okay. If you're not supposed to use logs, exponentials, functions or formulas, or guess-n-check, then what method were you supposed to use? What steps did your fellow students use (the ones who got the points for this)?

Please be complete. Thank you! ;)
 
Our test was about percentage. What my classmates used I cant remember, but if im correct, it was something similiar to equation.
 
Zulgok said:
Our test was about percentage.
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Perhaps it was about percentages? And perhaps this equation, as was pointed out earlier, relates to the formula for growth of an investment given a particular interest-rate percentage?

Zulgok said:
What my classmates used I cant remember, but if im correct, it was something similiar to equation.
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It (the solution method?) was similar to what equation? Please be specific. Thank you! ;)
 
Ishuda described this perfectly : 'How many years(x) does it take $4000 to grow to $4622.50 at a compounded 7.5% interest per year'.
 
The question looks like a 'How many years does it take $4000 to grow to $4622.50 at a compounded 7.5% interest per year'. So, if you don't know logarithms, just start working, or as Subhotosh Khan said, use trial and error:
at the end of year 1: 4000*1.075=4300
at the end of year 2: 4300*1.075=...
Click to expand...
Zulgok said:
Ishuda described this perfectly : 'How many years(x) does it take $4000 to grow to $4622.50 at a compounded 7.5% interest per year'.
Click to expand...


So what did you get following his suggestion?
 
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