All Questions
Tagged with algebra-precalculus exponential-function
692
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Question on exponential functions and drawing suitable lines on the graph
This question is from an exercise in an Edexcel Further Pure Mathematics book. Parts a, b, and c were fairly easy for me. For part a, I found the corresponding values of y using a calculator and in ...
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What to consider when taking kth root on both sides of equality
Say I have the following expression:
$10^{l} = a^{k}$
If I take the kth root of both sides, does that mean we get:
$10^{\frac{l}{k}} = a$
We don't have to consider anything with plus or minus?
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How to account for domain when solving inequalities in exponential functions
I'm struggling with the following question from an algebra exam from the 1970s.
The function f is defined as follows: $f(x) = \frac{3^{x+1} + 1}{3^x-1}$.
a) Solve the inequality $f(x) < 5$
b) For ...
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Isolating $z$ in the equation $x - 1 = - \frac{1-y^{z+1}-0.5(1-y^z)}{(1-y)y^z}$ [closed]
I have a formula with multiple unknowns:
$$x - 1 = - \frac{1-y^{z+1}-0.5(1-y^z)}{(1-y)y^z}$$
The way it is setup now allows to easily calculate $x$, but I would like to reformulate it to isolate $z$, ...
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I need help filling in some in a step from Fomin's calculus of variations
At the bottom of page 20 from Fomin's book on Calculus of Variations, we have:
(1) $\frac{x+A}{c}= \ln( \frac{y + (y^2-c^2)^{1/2}}{c})$
Implies that
$y = c \cosh(\frac{x+a}{c})$
Can somebody help me ...
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$(-a)^x$ versus $-(a^x)$ help
$(-2)^3=-8$ and $(-2)^2=4$, right? And $-(2^3)=-8$ and $-(2^2)=-4$.
So that means $(-a)^x$ does not equal $-(a^x)$.
My question is why do we never see graphs of $(-a)^x$ then?? I tried graphing $(-2)^...
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Why is $\sum_{m=0}^{\lfloor xs\rfloor} 2 \binom{s}{m} p^m (1-p)^{s-m} \leq 2\exp{\left(-\frac{2(\lfloor xs\rfloor - sp)^2}{s}\right)}$
I am trying to understand few of the mathematical steps I have encountered in a paper, there are two of them
(a) $\sum_{m=0}^{\lfloor xs\rfloor} 2 \binom{s}{m} p^m (1-p)^{s-m} \leq 2\exp{\left(-\frac{...
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how do you find the initial amount of a decay problem when you dont have one in the problem? [closed]
Suppose a sample of a certain substance decayed to $65.2\%$ of its original amount after $300$ days.
What is the half-life (in days) of this substance? (Round your answers to two decimal places.)
I ...
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Proof using the Lambert W function that 1 = 0 - What went wrong?
All values that satisfy $x^2=2^x$ would satisfy $\ln(2)x^3 = x\ln(2)e^{x\ln(2)}$, and would therefore satisfy the relationship $W(\ln(2)x^3) = x\ln(2)$. The problem is that when I graph these ...
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An analytic solution to solve $x^9=3^x$
I want to find a way to solve $x^9=3^x$ analytically, for two roots. one of them can be found below $$x^9=3^x\\(x^9)^{\dfrac {1}{9x}}=(3^x)^{\dfrac {1}{9x}}\\x^ { \ \frac 1x}=3^{ \ \frac 19}\\x^ { \ \...
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Why does $y = \frac{2A\sin(x\pi \ell)}{\pi^2 \ell (1-\ell) x^2}$ simplify to $y=c/x$ as $\ell$ approaches $0$?
I am working on something using this equation, and I find something strange. I am manipulating $\ell$ here between $0$ and $1$. I note that as $\ell$ approaches zero ($\sim0.001$ or less) it becomes a ...
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Solutions to Some Logarithmic Inequalities
Suppose we have an inequation as shown below:$$I_0:\space \ln (x) > \frac{x-2}{x}$$ Now we would like to find the largest set $S$ of real numbers such that any element $p\in S$ will satisfy $I_0$ ...
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How to evaluate $(1-aP)^P=d$ for P where $P>0$ and a & d are known
For context I am working trying to work out a formula based on the payout annuity formulas.
With these forums I have been able to derive the expected term for a given payment:
Standard payout annuity ...
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Show that for $a \neq b$ it holds: $\frac{e^b-e^a}{b-a} < \frac{e^b+e^a}{2}$ [duplicate]
Show that for $a \neq b$ it holds:
$$\frac{e^b-e^a}{b-a} < \frac{e^b+e^a}{2}$$
My first idea was to rearrange
$$2 \cdot (e^b-e^a) < (b-a)(e^b+e^a)$$
$$2e^b-2e^a < be^b + be^a - ae^b -e^a$$
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Solving $\ln\left(\frac{1}{x-2}\right)=\frac{1+2e^x}{e^x}$ [closed]
Here's the question I came across, they're inverses in this case, but I imagine that there is a way to do that without them being inverses.
$$\ln\left(\frac{1}{x-2}\right)=\frac{1+2e^x}{e^x}$$
