# How to get around an overflow error in Ti-84 plus calculator?

so i'm having a problem where i'm trying to find the answer to this equation

Ŷ = (10^-509.0738)(2020^158.0975)

The problem is that when i plug it into my calculator, it says an overflow error. Is there anyway to get around this or is there an alternative way of solving it?

Thanks.

### 3 Answers

- ?Lv 79 years agoFavorite Answer
Yeah, the TI-84 Plus can only handle a maximum exponent of 99. Anything over that will get you Err:Overflow.

Given:

y = (10^-509.0738)(2020^158.0975)

Take the appropriate logs:

log y = log(10^-509.0738)(2020^158.0975))

log y = log(10^-509.0738) + log(2020^158.0975)

log y = -509.0738 + 158.0975log(2020)

log y = -509.0738 + 522.5677881

log y = -509.0738 + 522.5677881

log y = 13.49

y = 10^13.49

y ≈ 3.1188 × 10^13

- husoskiLv 79 years ago
Yes. Use base 10 logarithms, plus some pencil and paper (or memory registers).

log Ŷ = -509.0738 + (log 2020)*158.0975

= 13.493988

The integer part of your answer is the power of 10 of the answer in scientific notation, and the decimal fraction is the base-10 log of the significant digits.

Ŷ = 3.118804 x 10^13

That works because 13.493988 = 13 + 0.493988

If you got a negative answer, like -13.493988, then this is not -13 + 0.493988. Instead you have

-13 - 0.493988 = -14 + 1 - 0.493988 = -14 + 0.506012

1/Ŷ is then 3.206358 x 10^-14

In this problem, the integer part was small enough that you could just use your 10^x or x^y function on the calculator. In other problems, the integer part might be too large to represent, so you'd manually separate the integer and fraction parts of the logarithm to get a pencil-and-paper answer in scientific notation.

- difioraLv 43 years ago
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