can someone show me these 2 problems on the calculator?

[harley4me2]

Log3.2x10^-5=

-log1x10^-5=

 

[skeeter]

1st question ... what kind of calculator you using?

2nd question ... from your post Log3.2x10^-5=, do you mean

\(\displaystyle \L \log{(3.2)} \times 10^{-5}\) or \(\displaystyle \L \log{(3.2 \times 10^{-5})}\), or something else?

same question would apply to the other expression you posted ...
-log1x10^-5=

 

[soroban]

Hello, harley4me2!

I hope you're familiar with most keys on your calculator.

To enter a negative number, do not use \(\displaystyle \fbox{\,-\,}\) (subtraction).
. . You should have a key that looks like this: \(\displaystyle \fbox{(-)}\)

On most calculators, when you press \(\displaystyle \fbox{\text{log}}\), it shows: \(\displaystyle \:\text{\log(}\)
. . automatically providing a left parenthesis.
On most calculators, you don't have to enter a right parenthesis \(\displaystyle \fbox{)}\) to close it.

\(\displaystyle \log\left(3.2\,\times\,10^{-5}\right)\)

Enter: \(\displaystyle \:\fbox{\text{\log}} \; \fbox{3.2} \; \fbox{\times} \; \fbox{10} \:\fbox{\wedge} \; \fbox{(-)} \;\fbox{5} \; \fbox{=}\)

. . You should get: \(\displaystyle \,-4.494850022\)

\(\displaystyle \text{-}\log\left(1\,\times\,10^{-5}\right)\)

Is that really "one times" in there? . . . We can ignore it!

We have: \(\displaystyle \:\text{-}\log\left(10^{-5}\right)\)

Enter: \(\displaystyle \:\fbox{(-)} \; \fbox{\text{log}} \;\fbox{10} \; \fbox{\wedge} \; \fbox{(-)} \; \fbox{5} \; \fbox{=}\)

. . You should get: \(\displaystyle 5\)

 

[harley4me2]

Thank you so much Soroban! You have helped me a ton!