can 2e^17 be (2e)^17? In other words do I 2 * e first then ^17 or do ^17 first then *2?
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can 2e^17 be (2e)^17? In other words do I 2 * e first then ^17 or do ^17 first then *2?
(2*e)^17 = (2^17)*(e^17), which is not 2*e^17
I want to reassure:
2e^17 is not (2*e)^17, correct?
BTW,
Did I do it right?Code:5(2)^3x - 4 = 13
10 ^ 3x = 17
log10 10 ^ 3x = log10 17
x = 1/3 log10 17
Exponention has priority over multiplication so 5(2)^3x <> 10^3x
ANY BODy?
>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)?
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.
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