Ln deducing

C

curicuri

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May 30, 2019
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Hello, have someone time for helping me out?

Question
I dont get that these expressions are the same, could someone help me to understand that?
12398
 
\(\displaystyle e^{ab}=\left(e^a\right)^b\)

\(\displaystyle e^{\ln(10)y}=\left(e^{\ln(10)}\right)^y\)

can you finish?
 
Thank you for reaching out so quickly!
Before continuing. I dont find a tool for symbols and math expressions. Are you using such thing? Or do you perhaps "copy & pasting"from other places?
 
this forum supports Latex. Basically anything enclosed in [mex][/mex] with the m changed to t (if there's a simple way to get the delimiters to display I didn't find it :p ) is interpreted as a LaTex expression and rendered as such.

Search the forums on Latex and you'll find much more information.
 
Last edited:
Romsek said:
... if there's a simple way to get the delimiters to display I didn't find it ...
Click to expand...
Hi Romsek. We can suppress BBCodes from rendering, by using [plain] and [/plain] tags (i.e., the new version of 'noparse'). They're explained in this thread. Cheers

?
 
Romsek:

I am not sure, perhaps eliminating e
ln(e^yln10)
yln10

If you dont have time it is also ok if you refer me somwehere about this stuff so that I can learn more about it....
 
you don't know what \(\displaystyle e^{\ln(10)}\) is?

perhaps review the relationship of the natural log with the exponential function
 
Ok, so I learned about the inverse relationship between e and ln.
e^(ln10)y=10^y
Super thanks to you Romsek!?
 
curicuri said:
Ok, so I learned about the inverse relationship between e and ln.
e^(ln10)y=10^y
Super thanks to you Romsek!?
Click to expand...

No, those two are not equivalent.

e^(ln10) is a constant and y is a variable.

e^(ln10)y = 10y

Because of how the Order of Operations works, you will need
more grouping symbols to keep the characters in that same
order:

e^[(ln10)y] = 10^y
 
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