Questions tagged [logarithms]

Questions related to real and complex logarithms.

The logarithm is generally defined to be an inverse function for the exponential. If $x > 0$ is a real number and $b > 0$, $b \ne 1$, then the base-$b$ logarithm is defined by

$$\log_b(x) = y \iff b^y = x$$

The most commonly used bases are base $10$ and $2$ (which frequently arises in computer science), and particularly base $e$. The natural logarithm $\ln$ is defined to be $\log_e$.

Alternatively, the natural logarithm can be defined to be a primitive of the function $$f(t) = \frac{1}{t}$$ subject to the condition that $\ln{1} = 0$.

In the study of complex numbers, the solutions $a$ of $e^{a} = z$ are called complex logarithms. This uniquely specifies the modulus of $a$, but not its argument; as such, we define the principal logarithm $\operatorname{Log}(re^{i\theta}) = \ln{r} + i \theta$, with the restriction $-\pi < \theta \le \pi$ (or alternatively, $0 \le \theta < 2\pi$). This leads to a branch cut, or discontinuity - alternatively, the complex logarithm can be viewed as a multi-valued function.

Reference: Logarithm.

This tag often goes along with .

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Solve $4^x \cdot 5^{4x+3} = 10^{2x+3}$ for $x$

I've got an equation I need to solve for $x$ but for the life of me can't work out how the answer is $$x = \log_{25}8 $$ The original equation is $$4^x \cdot 5^{4x+3} = 10^{2x+3}$$ Could someone please point me in the right direction?
David E
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Characterstic of log

I could not understand how they got $T_{100}$ as 300 . Can anybody please explain me .
Koolman
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Can someone help in reversing this equation?

I have Tf1(ADC1), but need to find out ADC1
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Solving Log(1.66) without a calculator.

I am taking a math class online in which I need to solve Algebra problems by hand without a calculator. I am getting caught up on how to solve log by hand. Whats a simple way to solve log(1.66)? we have to test via webcam with a piece of paper, and…
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Simplifying $2^{\sqrt {2\lg t}}$

Simplifying $2^{\sqrt {2\lg t}}$ should yield $t^{\sqrt {2/\lg t}}$, but the following is what I am getting: $$\begin{align} y &= 2^{\sqrt {2\lg t}}\\ y &= 2^{\sqrt {\lg t{^2}}}\\ y &= 2^{({{\lg t{^2}})}^{1/2}}\\ \lg y &= \lg{2^{({{\lg…
ssss
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Applying Natural log to an equation

(Note: I'm preparing for an Professional Engineering exam on topics I have not used in over 30 years. So questions might be overly basic. My apologies.) 1) If I take the natural log of one side of an equation, do I have to do so with the other side…
Glenn
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Find $\log_{256}N$ if $N=(2+1)(2^2+1)(2^4+1)\cdots(2^{32}+1)+1$

Find $\log_{256}N$ if $N=(2+1)(2^2+1)(2^4+1)\cdots(2^{32}+1)+1$. I can't think of any way to begin solving this question. Can someone help me with this question? A hint to help me obtain the right approach to solve this question would…
oshhh
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Calculator gives wrong answers for log

$10^3 = 1000$ My calculator with $10\log(1000)$ gives $30$ $15^4 = 50625$. My calculator with $15\log(50625)$ gives $70.57$. What am I doing wrong? Or would there be something wrong with my properties? I am using a Casio fx-82MS. Thanks in…
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Natural logarithm simplification

$$\begin{align} u(x_1x_2) & = -a\ln{a\over (a+b)x_1} -b\ln {b\over (a+b)x_2} \\ & = a\ln x_1 + b\ln x_2 + \text{constant}. \end{align}$$ Could somebody please explain how this is simplified? I'm confused with the steps needed to give us these…
metrix
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Solving for n involving logarithm

I know how to solve if the left side is just lgn, but it has an extra n. Was it correct to multiply both sides by e?
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Logarithm equation (solving for $x$)

I am stuck on this problem. Find the value of $x$ in this expression: $$ \ln(x-2) \cdot \ln(x)=\ln(64) $$
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Solve logarithmic inequality

I need help solving $-\log_2(x-1)*\log_2(3x-4)>0$ The logarithms have a base number of $2$.
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Logarithms question: $6+\log_3 (y) =\log_3(y)^5$

Logarithms question: $6+\log_3 (y) =\log_3(y)^5$ Forms part of a simultaneous equation. FULL QUESTION: Solve for $x$ and $y$: \begin{cases} \log_3(xy) =5 \\[4px] \log_3(x)\log_3(y)=6 \end{cases}
Cicada
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Solving the inequality $2^n <2^{2n}-1000$

I am trying to solve the inequality $2^n <2^{2n}-1000$ using purely elementary properties of the logarithm to obtain an exact solution. With some re-arranging and then taking the logarithm base $2$ of both sides I get: $n+\log_2(2^n-1)>\log_2(1000)$…
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Define $g(x)=\log_b(x)$ for $b>0$ in terms of the exponential function $f(x)=b^x$.

I know that these two functions are inverses of each other, therefore how can you define one in terms of its inverse?
Chad
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