What is the solution of log2x − 5 25 = 2?
i don't quite understand the notation....is the 2 between the log and the x small? is that 525, or something else?
assuming you want to solve for x:| log(2x)=2+525 2x=e^(527) x=[e^(527)]/2
agree with @mtbender74 the notation is cryptic
it would be 10 not e @VeritasVosLiberabit ...common log not ln
@mtbender74 that is again a notation problem log can be either base 10 or base e. It depends on how its defined
assuming, of course, that's what the question really is...and not a log base 2
that could also be the case, we'll just have to see what @princesssarah says
for all we know it could be \[log_{2}(x-5)^{25}=2\]
The drawing tool or equation tool may help. That way everyone will know if you mean \(\log_2x\) or \(\log (2x)\) or \(log2^x\)
etc.
did this question few minutes back, its : \(\large \log_{2x-5} 25 = 2\)
change it into exponent form and compare bases
ahhh...that's much nicer :)
oh well there you go
wait what
\(\large \log_a b = c \iff b = a^c\)
\(\large \log_{2x-5} 25 = 2 \iff 25 = (2x-5)^2\)
\[(2x-5)^{\log _{2x-5}25}=(2x-5)^{2}\]
and use the fact that 25 = 5^2
\(\large \log_{2x-5} 25 = 2 \iff 25 = (2x-5)^2\) \(\large 5^2 = (2x-5)^2\)
so its 5
since the exponents are equal, u can safely compare the bases : \(\large 5 = 2x - 5\)
yes ! solving for x gives u x = 5
For the graphed function f(x) = (4)x - 1 + 2, calculate the average rate of change from x = 2 to x = 4.
I don't know how to put the picture
do u have the picture already ?
if so, just click "Attach File" right below this textbox
\(\large y = 4^{x-1}+2\)
start by finding the value of y at x = 2 and x = 4
\(\large y = 4^{x-1}+2\) plugin x = 2, what do u get ?
6 and 66
Excellent ! use the formula for average rate of change : \(\large \dfrac{f(4)-f(2)}{4-2} = \dfrac{66-6}{2}\) simplify
30
Yep ! hey next question please post in a separate thread... that way others get to see and answer quickly...
you may "Close this quesiton" u will see that button right below ur question :) @beccaboo333 please guide the new user if you're here :)
@uri
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