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Mathematics 16 Online
OpenStudy (anonymous):

ln 10 + lnx = 0????

OpenStudy (lalaly):

ln 10x= 0 10x = e^0 10x=1 x=1/10

OpenStudy (anonymous):

where did the e come from?

OpenStudy (anonymous):

i know ln=e but.. im confused.

OpenStudy (lalaly):

The natural logarithm function ln, is the inverse function of the exponential function, leading to the identities: \[\large{e^{lnx} = x}\]\[\ln(e^x)=x\]

OpenStudy (lalaly):

and ln(e) =1

OpenStudy (anonymous):

e is the exponent of nature log

OpenStudy (lalaly):

so first ln10+lnx = ln(10x) rules of logarithms ln10x = 0 to get rid of the ln \[\huge{e^{\ln10x}} = e^0\]

OpenStudy (lalaly):

\[\huge{\cancel{e}^{\cancel{\ln}10x}} = e^0\]

OpenStudy (anonymous):

e=1+1/1!+1/2!+1/3!+1/4!+.....

OpenStudy (lalaly):

\[\huge{10x=1}\]

OpenStudy (lalaly):

\[\huge{x=\frac{1}{10}}\]

OpenStudy (anonymous):

ln(10x) = 0?

OpenStudy (lalaly):

yeah

OpenStudy (anonymous):

anything power to 0 equal to 1

OpenStudy (lalaly):

:)

OpenStudy (anonymous):

^^

OpenStudy (anonymous):

I feel extra dumb.. Imma have to write this down. Bu the part where you said the identities can you say it in english?

OpenStudy (lalaly):

which part? my second post?

OpenStudy (anonymous):

Ya.

OpenStudy (anonymous):

so when i was trying to get rid of the ln on the left side why did i add an e on both sides instead of an ln?

OpenStudy (lalaly):

u add e so u can solve for x only because e is the inverse of ln so they cancel

OpenStudy (anonymous):

so whats e^2?

OpenStudy (lalaly):

e^2? why e^2

OpenStudy (anonymous):

cause e^0 = 1? right? so whats the rule of it?

OpenStudy (lalaly):

any number raised to the power 0 is 1... e^2 is just some other number, u can find it on the calculator

OpenStudy (anonymous):

okay so , ln(e) x = x ='s e^x=x?

OpenStudy (anonymous):

i cant see the full link?

OpenStudy (anonymous):

\[\ln x = \log_{e} x\]Also\[\log_{a}b = \frac{\log b}{\log a}\]So\[\ln x = \log_{e}x = \frac{\log_{10} x}{\log_{10} e} = \frac{\log_{10}x}{0.43429...}\]So you can convert betwen ln(x) and log(x) by just multiplying by a constant.

OpenStudy (anonymous):

if I were a number then rashan^0=1 any number raised to 0 is 1. That means: e^0=1

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