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Question about changing a logarithm's base
I've been using the following method to derive/remember the logarithm base conversion formula: If I want to convert $\log_{a} (x)$ to an expression in base b, I say,
$a^{\log_{a} (x)} = x$
$\log_{b} (a^{\log_{a} (x)}) = \log_{b} (x)$
$\log_{a} (x) \log_{b} (a) = \log_{b} (x)$
$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$
It kind of feels like I'm working backwards, and I was wondering if there's a more direct way to go about it. I tried:
$\log_{a} (x) = \log_{a} (b^{\log_{b} (x)})$
$\log_{a} (x) = \log_{b} (x) \log_{a} (b)$
but couldn't rid myself of the loga in the right-hand side of the equation. It occurred to me that if I could rewrite the "a" subscript as " $b^{\log b (a)}$ ", I might be onto something. (Or might not.) Does the notation ever get used like that, where you perform a substitution in a subscript?

Dumaen80p3 2022-03-30

Prove that if $e < y < x$ then $x^{y} < y^{x}$
My try:
I tried to use Taylor's theorem:
$y^{x} - x^{y} = e^{x \ln y} - e^{y \ln x} = 1 + x \ln y + o (x \ln y) - 1 - y \ln x - o (y \ln x) =$
$= {\ln y}^{x} - {\ln x}^{y} + o (x \ln y) - o (y \ln x) = \ln \frac{y^{x}}{x^{y}} + o (x \ln y) - o (y \ln x)$
Hovewer I have a problem to show that $y^{x} - x^{y} > 0$ because I can't say that $\ln \frac{y^{x}}{x^{y}} > 0$ because then I use with theses.
Have you some idea?

metalskaashw 2022-03-29

Prove the inequality $e^{x} \geq x^{e}$ for $x > 0$

Sage Fry 2022-03-28

Consider the exponential equation. $9^{x} = 733$
a. Find x in exact form: $x = b$ . Approximate x, correct to at least 3 decimal places: $x =$
b. Approximate $x$ , correct to at least 3 decimal places: $x =$

Janiyah Hays 2022-03-28

$10^{- 9} [2 \times 10^{6} + 3^{1000}]$
I'm stuck on solving this. I wasn't able to put this into my calculator since the number is too big for it to calculate.
So far I've done this:
$2 \times 10^{- 3} + 3^{1000} \times 10^{- 9}$
log the equation
$\log [2 \times 10^{- 3} + 3^{1000} \times 10^{- 9}]$
$\log (2 \times 10^{- 3}) + \log (3^{1000} \times 10^{- 9})$
$(\log 2 + {\log 10}^{- 3}) + ({\log 3}^{1000} + {\log 10}^{- 9})$
$(\log 2 - 3) + (1000 \times \log 3 - 9)$
How do I get rid of the log afterwards?

monkeyman130yb 2022-03-28

$\ln r + \ln q = k r$ Isolating $r$

rhedynogh0rp 2022-03-28

$\int \frac{f}{f^{'}} d x = \int \frac{1}{{(\ln f)}^{'}} d x = ?$
I recently saw in another post that ${(\ln f)}^{'} = \frac{f^{'}}{f}$
where $f = f (x)$
From which it follows that $\int \frac{f^{'}}{f} d x = \int {(\ln f)}^{'} d x = \ln f + C$
What about integrating the inverse of this?
I.e. what about $\int \frac{f}{f^{'}} d x = \int \frac{1}{{(\ln f)}^{'}} d x = ?$

Jaylin Clements 2022-03-28

Proving that $\sum_{{n = 1}}^{\infty} \frac{{(- 1)}^{n + 1}}{n} = \log 2$

metalskaashw 2022-03-27

difference of logs
$\log (a - b) - \log (a - c)$
oes this have a simpler form? Perhaps one where the as have cancelled out? I know it can also be expressed as a log of the fraction: $\log \frac{(a - b)}{(a - c)}$ , but the same question applies.

Guadalupe Glass 2022-03-27

Continuing logarithm $\log (\log (\dots \log (z)))$

Dumaen80p3 2022-03-27

How can i solve this $4^{2 + x} + 15 \cdot 4^{x^{2}} - 16^{x^{2} - \frac{x}{2}} = 0 ?$

Pasegeabe85xy 2022-03-27

How can I solve the following equation:
$2^{\log_{3} x} + x^{\log_{3} 2} = 4$
I don't want the final answer, I want to know how I can solve these kind of equations.

ropowiec2gkc 2022-03-26

Derivative of $x^{x}$ without logarithmic differentiation
With logarithmic differentiation, it is quite simple to compute the derivative of $x^{x}$ :
$y = x^{x}$
$\ln y = x \ln x$
$\frac{1}{y} \frac{d y}{d x} = \ln {x} + 1$
$\frac{d y}{d x} = (\ln (x) + 1) x^{x} .$
Is there a method to compute the derivative of $x^{x}$ that does not rely on logarithmic differentiation?

Nancy Richmond 2022-03-26

Derivative for log
I have the following problem:
$\log (\frac{x + 3}{4 - x})$
I need to graph the following function so I will need a starting point, roots, zeros, stationary points, inflection points and local minimum and maximum and I need to know where the function grows and declines.
I calculated roots zeros $x + 3 = 0, x = - 3$ and roots $4 - x = 0, x = - 4$ . Now I sort of know how to graph the function from here but how do I get the stationary points do I have to find the derivative of $\log (\frac{x + 3}{4 - x})$ or just $(\frac{x + 3}{4 - x})$
I don't fully understand how to find the derivative of log. Can i use the ${\log (x)}^{'} = \frac{1}{x}$ rule here to get $\frac{1}{\frac{x + 3}{4 - x}}$ and then find stationary points here ?

ideklaraz7xz 2022-03-26

Could you describe this function as "logarithmic"?
$f (x) = \frac{1}{\sqrt{x}}$
As x increases, the value of $f (x)$ decreases, but the decrease tapers off quickly as $x$ gets larger, and if you plot the graph of $f (x)$ , the shape looks kind of like an upside-down logarithm. Would it be correct to describe this function as declining logarithmically, as $x$ increases?

Kason Palmer 2022-03-26

How can we solve:
$\sqrt{x} + \ln (x) - 1 = 0 ?$

Lesnaoq73 2022-03-26

How can we interpret the difference between two log points? Is it correct to interpret this difference in percentage points?

Lorelei Stanton 2022-03-26

How can I show that $lim_{x \to \infty} \frac{\log (x + 1)}{\log x} = 1 ?$

Oliver Carson 2022-03-26

How can I calculate $lim_{x \to 0} \frac{\log (\cos (x))}{\log (\cos (3 x))}$ without l'Hopital?

amonitas3zeb 2022-03-25

Determine if the equation is valid/true
The equation is:
$\log_{b} \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} = 2 \log_{b} (\sqrt{3} + \sqrt{2}) .$
I can get as far as:
$\log_{b} (\sqrt{3} + \sqrt{2}) - \log_{b} (\sqrt{3} - \sqrt{2}) = 2 \log_{b} (\sqrt{3} + \sqrt{2})$
Which looks almost too simple, but I can't get the signs to match up right to solve the problem. Do I need to further break out the logarithmic functions that are there?

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