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How many numbers less than $x$ have a prime factor that is not $2$ or $3$
I am trying to figure out the number of integers greater than 1 and less than or equal to x that have a prime factor other than 2 or 3. For example, there are only two such integer less than or equal to 7.
It is straight forward to determine how many many integers less than or equal to x have a prime factor other than 2:
$x - ⌊ \log_{2} x ⌋$
Or to make the same determination about 3:
$x - ⌊ \log_{3} x ⌋$
What is the method or formula for figuring out how many integers less than or equal to x have a prime factor other than 2 or 3?
I know that it is less than:
$x - ⌊ \log_{2} x ⌋ - ⌊ \log_{3} x ⌋$
and greater than:
$x - ⌊ \log_{2} x ⌋ - ⌊ \log_{3} x ⌋ - ⌊ \frac{x}{6} ⌋$
Thanks

ajumbaretu 2022-11-18

Using logarithms to solve the following equation to find x
$9^{2 x} = 27^{1 - x}$ ?? I'm really struggling with this questions. I appreciate your help and if you can please show me your working out so I can understand it too

Cortez Clarke 2022-11-18

Solve logarithmic equation for $x$ to find the inverse of $f (x) = \ln (x + \sqrt{x^{2} + 1})$

Laila Murphy 2022-11-18

Find the solution to the differential equation
Assume $x > 0$ and let
$x (x + 1) \frac{d u}{d x} = u^{2},$
$u (1) = 4.$
I started off by doing some algebra to get:
$\frac{1}{u^{2}} d u = \frac{1}{x^{2} + x} d x .$
I then took the partial fraction of the right side of the equation:
$\frac{1}{u^{2}} d u = (\frac{1}{x} - \frac{1}{x + 1}) .$
I then took the integral of both sides:
$- \frac{1}{u} = \log x - \log (x + 1) + C .$
From here I don't know what to do because we are solving for $u (x)$ and I'm not sure how to get that from $- \frac{1}{u}$

Howard Nelson 2022-11-18

Dealing with Logarithms. $\log (b^{x} + a) = \log (c)$
What methods/techniques are available to solve for x in the following type of situation:
$\log (b^{x} + a) = \log (c)$
The only log methods I have been exposed to are using the power laws and bring x out, which you cannot do in this case.
Thanks for your help.

Melissa Walker 2022-11-18

Limit of logarithmic function using l'Hospital
How can I find the following limit:
$lim_{x \to \infty} \frac{\ln (1 + α x)}{\ln (\ln (1 + e^{β x}))}$
where $α, β \in R^{+}$
My first guess was to use l'Hospital:
$lim_{x \to \infty} \frac{\ln (1 + α x)}{\ln (\ln (1 + e^{β x}))} = lim_{x \to \infty} \frac{\ln (1 + e^{β x}) (1 + e^{β x}) α}{(1 + α x) e^{β x} β}$
But what can I do now? Is my approach correct or is there a simpler method?
Edit:
$lim_{x \to \infty} \frac{\ln (1 + e^{β x}) (1 + e^{β x}) α}{(1 + α x) e^{β x} β} = lim_{x \to \infty} \frac{\ln (1 + e^{β x}) α}{(1 + α x) β} lim_{x \to \infty} (1 + e^{- β x})$
Applying L'Hospital a second time on the first fraction,
$lim_{x \to \infty} \frac{\ln (1 + e^{β x}) α}{(1 + α x) β} lim_{x \to \infty} (1 + e^{- β x}) = lim_{x \to \infty} \frac{α β e^{β x}}{(1 + e^{β x}) α β} \cdot 1 = lim_{x \to \infty} \frac{e^{β x}}{1 + e^{β x}} \cdot 1$
Now let's apply L'Hospital one final time:
$lim_{x \to \infty} \frac{e^{β x}}{1 + e^{β x}} \cdot 1 = lim_{x \to \infty} \frac{e^{β x} β}{e^{β x} β} \cdot 1 = 1$
Is this correct?

clealtAfforcewug 2022-11-18

Integrate $\int (\log (\sin x \cos x))^{n} d x$ with hypergeometric function form
Could anyone help me with that?

Abdiel Mays 2022-11-18

Simplify a log expression
I met a problem, I don't know if this term can be simplified properly?
$e^{(\ln a x^{b})^{c}}$
since the ln term with power of c is hard to cope with, thanks for any help!

Hanna Webster 2022-11-18

How do I evaluate this without using taylor expansion : $lim_{x \to \infty} x^{2} \log (\frac{x + 1}{x}) - x$ ?
How do I evaluate this without using Taylor expansion?
$lim_{x \to \infty} x^{2} \log (\frac{x + 1}{x}) - x$
Note: I used Taylor expansion at $z = 0$ and I have got $\frac{- 1}{2}$
Thank you for any help

apopiw83 2022-11-18

Use properties of lagrithma to evaluate without calculator
$2 \log_{9} (3) - 4 \log_{9} (3) + \log_{9} (\frac{1}{729})$

charmbraqdy 2022-11-18

How do I prove that $f (x) = l n (x) - (x - 4)^{2}$ has exactly two roots.
I assume we take the derivative of the function. I get: $y^{'} = 1 / x - 2 (x - 4)$ and I attempt to set it to 0 and solve but get stuck. Any tips?

Ty Moore 2022-11-18

Logarithm problem
If $a^{x} = b^{y}$ , then how come $x \log a = y \log b$ holds? Can anyone show me how this is with all steps and necessary logarithm formula?

paratusojitos0yx 2022-11-17

Limit evaluate $lim_{x \to 0} \frac{\ln (\cos (- 5 x))}{\ln (\cos (- 3 x))}$
Please help me with this limit without using L'Hôpital's rule. I would by happy if you use simple solving. Thank you as much as I can ;).

Uroskopieulm 2022-11-17

Solving logarithmic equation, different bases
What number do I need to multiply both sides with? I have worked for an hour on this but it is the first time I am using this website so it is impossible for me to write what I have already done. If you can give me tips on how to solve it I would appreciate it a lot!
$2 \log_{6} (\sqrt{x} + \sqrt[4]{x}) = \log_{4} x$

bruinhemd3ji 2022-11-17

Infinite inequality with logarithm
Solve inequality $3 - \log_{0.5} x - \log_{0.5}^{2} x - \log_{0.5}^{3} x - \dots \geq 4 \log_{0.5} x$ Any suggestions how to start?

Owen Mathis 2022-11-17

Log Log Integrals II
The integral
$\begin{aligned} I_{4} = \int_{0}^{1} \ln (1 - x) \ln^{2} (\ln (\frac{1}{x})) \frac{d x}{x} \end{aligned}$
can be expressed as
$\begin{aligned} I_{4} = ζ^{^{″}} (2) - \frac{γ^{2} π^{2}}{6} - \frac{γ π^{2}}{3} \ln (\frac{2 π}{A^{12}}) + \frac{π^{4}}{36} \end{aligned}$
where $A$ is the Glaisher-Kinkelin constant.
The integrals
$\begin{aligned} I_{5} = \int_{0}^{1} \frac{\ln (1 - x)}{\sqrt{\ln (\frac{1}{x})}} \ln^{2} (\ln (\frac{1}{x})) \frac{d x}{x} \end{aligned}$
and
$\begin{aligned} I_{6} = \int_{0}^{1} \frac{\ln (1 - x)}{\ln^{3 / 2} (\frac{1}{x})} \ln^{2} (\ln (\frac{1}{x})) \frac{d x}{x} \end{aligned}$
can be expressed in terms of $\partial_{s}^{2} ζ (s) |_{s = 3 / 2}$ and $\partial_{s}^{2} ζ (s) |_{s = 1 / 2}$ , respectively. Can these integrals be evaluated in closed form expression without the use of derivatives of the Zeta function?
If the integrals can be evaluated in such a way what is the resulting value?

Kailyn Hamilton 2022-11-17

How to solve this exponential equation? $2^{2 x} 3^{x} = 4^{3 x + 1}$
I haven't been able to find the correct answer to this exponential equation:
$\begin{aligned} 2^{2 x} 3^{x} & = 4^{3 x + 1} \\ 2^{2 x} 3^{x} & = 2^{2} \times 2^{x} \times 3^{x} \\ 4^{3 x + 1} & = 4^{3} \times 4^{x} \times 4 \\ 6^{x} \times 4 & = 4^{x} \times 256 \\ x \log_{6} 6 + \log_{6} 4 & = x \log_{6} 4 + \log_{6} 256 \\ x + \log_{6} 4 & = x \log_{6} 4 + \log_{6} 256 \\ x - x \log_{6} 4 & = \log_{6} 256 - \log_{6} 4 \\ x (1 - \log_{6} 4) & = \log_{6} 256 - \log_{6} 4 \\ x & = \frac{\log_{6} 256 - \log_{6} 4}{1 - \log_{6} 4} \\ x & = 10.257 \end{aligned}$
so when I checked the answer I wasn't able to make them equal, I have tried variants of this method but I feel I'm missing something..

kituoti126 2022-11-17

If $e^{2 π i} = 1$ then is $2 π i = 0$ ?
So with Euler identity I used $2 π (τ)$ instead and got $e^{2 π i} = 1$ , and took natural logarithm $2 π i = 0$ ? But is see online the answer is $6.28.... i$ .

Siemensueqw 2022-11-17

Solution of an exponential equation
Probably very simple question. Why the solution of
$1 = n (1 - a)^{t}$
in terms of t is equal to:
$t = \frac{\ln n}{\ln \frac{1}{1 - a}}$

Frankie Burnett 2022-11-17

Use logarithms to solve the equation for t.
$9 - e^{0.2 t} = 5$

Finding logarithms questions and answers online is always pure luck, especially if these are provided with detailed answers to questions. This is exactly what you can find below regardless of your practical application, college course, or complexity. These logarithms questions are related to higher education courses, which is why they will help you find the answers to what you currently have. When you deal with logarithmic equations you are approaching an exponential equation where the variable appears in an exponent as it’s in (log b (bx) = x. It also shows that the logarithmic graph equation can be represented as well.

Algebra II

Learn How to Graph Logarithmic Equations