anyone know what to do logarithms with a negative exponent on the TI-84 ?

Question

campbell_st · Answer

can you give an example

campbell_st · Answer

perhaps putting the negative exponent in brackets may help

anonymous · Answer

no I'm using an old schaum's outline series for College Algebra.  It's so old it still uses the tables with mantissas and characteristic.  I'm trying to do the same problem with my calculator

campbell_st · Answer

an an example 

$$\log(5^{(-4)}) = $$

campbell_st · Answer

and you'd get -2.796

anonymous · Answer

Their problem is Find N if if log N = 7.8657 - 10 ( the minus 10 makes it negative.  My ti-84 says answer is .000136623, the answer is supposed to be .00734

anonymous · Answer

sorry .00013623

anonymous · Answer

Minus 10 make it -3.8657

anonymous · Answer

I took 10^(-3.8657)

campbell_st · Answer

so you are working in base 10 logs.... 

so if you raise each side of the equation as a power of 10 you will get 

$$10^{logN} = 10^{7.8657 - 10}$$

which becomes 

$$N = 10^{(7.8657-10)}$$

so on your calculator enter 

10^(7.8657 - 10) then press equals

campbell_st · Answer

and you'll find the answer matches the text

anonymous · Answer

I did get same answer doing it your way and it matches the book, but why doesn't it work in ti when I actually take 10 to the -3.8657 power by actually subtracting 10 from 7.8657?

campbell_st · Answer

because it recognises - as an operation not part of a value... 

its like -2^2 = -4 on calculators     to get it right you need (-2)^2 =4

anonymous · Answer

but I did put it parentheses 10^(7.8657-10)   :(

campbell_st · Answer

put it down to a programming glitch

anonymous · Answer

Oh what an idiot I am! 7.8657-10 is not -3.8657, it's -2.1343

anonymous · Answer

fixing my subtraction gets it right on ti.  I will take your advice, assume it's  programing and enter exponent as (7.8657-10) .  Thanks