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kittybasil:
oops I spelled conversion wrong. But do you get what to do so far? @jknkjkjnjjjkl
jknkjkjnjjjkl:
log(b)/log7
kittybasil:
Actually, no - they explicitly mentioned the natural log, which is \(\large{log_{e}}\) or \(\ln\)
So instead we'd be calculating like this:\[\frac{\ln b}{\ln7}=?\]By the way, idk how many decimal places they want you to round to, so...
Florisalreadytaken:
@jknkjkjnjjjkl wrote:
log(b)/log7
look above -- what did i meantion about \( b=? \)
jknkjkjnjjjkl:
t
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Florisalreadytaken:
then plug that in!
kittybasil:
Oops, did it backwards. Flip that fraction...\[\frac{\ln7}{\ln t}=?\]This should be what you have.
kittybasil:
btw it's not log, it's ln or log e
jknkjkjnjjjkl:
n^6/b
kittybasil:
...wait what?
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Florisalreadytaken:
@florisalreadytaken wrote:
what your teacher wants you to do is, to do it in this way:
[color= inherit;][/color]\[\log _{a} b=\log b / \log a\]
where
[color= inherit;][/color]\[ a=7 \]
[color= inherit;][/color]\[ b=t \]
lapsus😂 -- i didnt notice in the second part -- its \( \ln \) instead of \( \log\)
Florisalreadytaken:
let me make it clear once again THE RIGHT WAY.
\[ \log _{a} b=\frac{\ln b }{ \ln a} \]
\[ a=7 \]
\
thus, we get this:
\[ \Large \log _{7} t=\frac{\ln t }{ \ln 7} \]