2e^17

**prog_tom** · Apr 6th, 2002, 11:12 AM

can 2e^17 be (2e)^17? In other words do I 2 * e first then ^17 or do ^17 first then *2?

**Guv** · Apr 6th, 2002, 11:25 AM

(2*e)^17 = (2^17)*(e^17), which is not 2*e^17

**prog_tom** · Apr 6th, 2002, 02:25 PM

I want to reassure:

2e^17 is not (2*e)^17, correct?

BTW,

Code:

5(2)^3x - 4 = 13

10 ^ 3x = 17

log10 10 ^ 3x = log10 17
x = 1/3 log10 17

Did I do it right?

Moon · Apr 6th, 2002, 02:59 PM

Exponention has priority over multiplication so 5(2)^3x <> 10^3x

**prog_tom** · Apr 6th, 2002, 03:32 PM

ANY BODy?

**prog_tom** · Apr 6th, 2002, 03:46 PM

>Logs are pretty perplex.

(3+4)^2, is it 3^2 + 4^2? or (7)^2.

What is (i+z+3)^2? I'm not talking about binomial theorem, is it (i^2 + z ^2 + 9)?

**Guv** · Apr 6th, 2002, 05:12 PM

2*e^17 = 48,309,905.5072

(2*e)^17 = 48,309,905.5072*2^16

(2*e)^17 = 3.166 037 967 34 E12

2*e^17 = (2)*(e^17)

(2*e)^17 = (2^17)*(e^17)

I hope there are no typo's in the above.

sql_lall · Apr 7th, 2002, 06:58 AM

When somthing is written in the form:
ax^b
This stands for a * (x^b), not (a*x) ^b

i.e.
5(2)^3x - 4 = 13
=>17= 5*(2^3x) (from above)
=>17/5 = 2^(3x)
=>log(17/5) = 3x * log(2)
=> log(17/5)/(3*log(2) = x

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