The solution to x = log91.43 is approximately 1.6. true or false?
can't u use a calculator to verify?
no :(
@iceywheniplease7 , what part are you having trouble with?
all of i dont understand the whole log solution thing and stuff
try this... \(\large 10^{1.6} \) if that gets you reasonably close to 91.43, then i'd say yes.... if not, then no...
Okay. Do you know what an exponent is?
yes
idt thats right.... haha
Well, a logarithm is basically the opposite of an exponent.
oh?
Well, besides their nasty name, they are really quite simple.
\(x^n=\{x*x*x\}\text{n-times}\) right?
Or, 4^13=4*4*4*4*4*4*4*4*4*4*4*4*4*4
A log is like the opposite of an exponent. If an exponent was multiplication, a log would mean division.
\(\log_44^{13}=13\)
or, \(\log_24^{13}=\log_22^{2^{13}}=\log_22^{26}=26\)
how bout \(\log_{10}13\)? well, we are basically asking this question. \(10^x=13\) Well, it turns out to be nasty. In fact, we can't do this by hand. Luckily, we can do it by calculator. I got something around 1.1139 If you try 10^(1.1139), you'll get a number close to 13. That is what a logarithm is.
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