ln 10 + lnx = 0????
ln 10x= 0 10x = e^0 10x=1 x=1/10
where did the e come from?
i know ln=e but.. im confused.
The natural logarithm function ln, is the inverse function of the exponential function, leading to the identities: \[\large{e^{lnx} = x}\]\[\ln(e^x)=x\]
and ln(e) =1
e is the exponent of nature log
so first ln10+lnx = ln(10x) rules of logarithms ln10x = 0 to get rid of the ln \[\huge{e^{\ln10x}} = e^0\]
\[\huge{\cancel{e}^{\cancel{\ln}10x}} = e^0\]
e=1+1/1!+1/2!+1/3!+1/4!+.....
\[\huge{10x=1}\]
\[\huge{x=\frac{1}{10}}\]
ln(10x) = 0?
yeah
anything power to 0 equal to 1
:)
^^
I feel extra dumb.. Imma have to write this down. Bu the part where you said the identities can you say it in english?
which part? my second post?
Ya.
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?
u add e so u can solve for x only because e is the inverse of ln so they cancel
so whats e^2?
e^2? why e^2
cause e^0 = 1? right? so whats the rule of it?
any number raised to the power 0 is 1... e^2 is just some other number, u can find it on the calculator
okay so , ln(e) x = x ='s e^x=x?
i cant see the full link?
\[\ln x = \log_{e} x\]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.
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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