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Mathematics 18 Online
OpenStudy (anonymous):

Log5(5) = x is x = 1? or 0?

OpenStudy (anonymous):

since \(5^1=5\) it is \(1\)

OpenStudy (anonymous):

I am having such issues with logarithms.

OpenStudy (anonymous):

if it was \(\log_5(1)\) then since \(5^0=1\) the answer would be \(0\)

OpenStudy (anonymous):

they are exponents

OpenStudy (anonymous):

this is what you need to be able to do quickly: go from \(\log_b(x)=y\) to \(b^y=x\) with no agony

OpenStudy (anonymous):

for example \[10^3=1000\iff \log_{10}(1000)=3\]

OpenStudy (anonymous):

or \(2^4=16\iff \log_2(16)=4\)

OpenStudy (anonymous):

I see. but what about when you start putting in e? For example: lne(8) = x

OpenStudy (anonymous):

\[\log_e(8)=x\iff x=e^8\]

OpenStudy (anonymous):

unless the question was \(\ln(e^8)\) which is a different story

OpenStudy (anonymous):

It was the first one

OpenStudy (anonymous):

then it is \(x=e^8\) if you want a decimal, you need to use a calculator

OpenStudy (anonymous):

2980.96?

OpenStudy (anonymous):

that is what i get too \[\log_{\heartsuit}(\spadesuit)=\clubsuit\] is is the same as \[\heartsuit^{\clubsuit}=\spadesuit\]

OpenStudy (anonymous):

is that all, or are there more?

OpenStudy (anonymous):

Idk, i'm still confused..

OpenStudy (anonymous):

suppose i am asked for \(\log_5(125)\) my job is to try to fill in the diamond \(5^{\diamondsuit}=125\)

OpenStudy (anonymous):

since i know \(5^3=125\) then i know \(\log_5(125)=3\)

OpenStudy (anonymous):

Okay... Is there like a method to find them other than memoriation? Because I was ever given like a list to remember XD

OpenStudy (anonymous):

it is not really memorization, it is more like knowing what it means

OpenStudy (anonymous):

if you know \(7^2=49\) then you also know \(\log_7(49)=2\)

OpenStudy (anonymous):

How do you know decimals though?

OpenStudy (anonymous):

i don't i would need a calculator

OpenStudy (anonymous):

i know \(\log_{10}(100)=2\) because i know \(10^2=100\) but if i was asked for \(\log_{10}(200)\) i would need a calculator to find it

OpenStudy (anonymous):

Oh

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