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 .

10168 questions
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How to see that $1$ is a solution of $x^{x^2−3x} = x^2$

I tried to solve the problem below to get all the positive solutions: $$x^{x^2−3x} = x^2$$ By using $\ln$ on both sides, I get that one solution is $\displaystyle\frac{3 + \sqrt{17}}{2}$. But $1$ is also a solution that you can guess. How can I see…
Dovendyr
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Prove that $a^x - b^y = 0$ where $x = \sqrt{\log_a b}$ & $y = \sqrt{\log_b a}$ , $a > 0$, $b > 0$ & $a, b \ne1$

I'm solving logarithm questions. I got stuck in this question. Prove that $a^x - b^y = 0$ where $x = \sqrt{\log_a b}$ & $y = \sqrt{\log_b a}$ , $a > 0$, $b > 0$ & $a, b \ne1$ I've tried to solve it. I'm unable to understand how to break that square…
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Solve logarithmic equation $2^x = 1 + 15\log_{5}(x+1)$

$\ 2^x = 1 + 15\log_{5}(x+1)$ Is there any other way of solving this equation, except graphical?
muffin
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How do I solve this logarithm equation with different bases?

How do is solve this logarithm equation? $$11 \cdot \log_3x+7 \cdot \log_7x = 13+3 \cdot \log_4x$$ I know that I have to use the change of base formula, but I still can't figure out the equation. Can someone help me?
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If $\log(ax)\log(bx) +1=0$ has a solution $x>0$, then find bounds on $b/a$

If equation $$\log(ax)\log(bx) +1=0$$ with constants $\;a>0,\; b>0\;$ has a solution $x>0$, it follows that $$\frac{b}{a} \ge ???$$ or $$???\ge\frac{b}{a}\gt???$$ Fill all in the blank. To be honest, I am very lost here and not sure how I can…
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Unknown in logarithm base $122312/100000 = (1+t)^5$

I am new to exponents and logarithms, and have been stuck with this for a quite long time. The problem is: $$\frac{122312}{100000}=(1+t)^5$$ or $$\log_{1+t}\frac{122312}{100000}=5$$ I have no idea, how to solve this. Hints or tips would be very very…
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Can we simplify this logarithm? if so, please provides some tips

${|x|^{11/10}} \log_{|x|^{{1/10}}}|x|$. I only know doing the first step, not sure if it is correct $\log_{|x|^{{1/10}}}(|x|^{|x|^{11/10}})$ as got stuck following this proof. Please help understand how we can get step two from step one.
Maxfield
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Solving continuous logarithm whith unknown base and unknown exponent

I was wondering if there is any feasible solution for finding the base and exponent of a continuous logarithm when only the result is known. Thus: $v=\log_{b}e$ with only $v$ given and $b,e$ are integers greater 1. By feasible I mean an approach…
Alex
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Let $n$ be the product of all positive factors of $10^4$. Find $log_{10} n$

I can't see any relationship between the products of the positive factors of powers of 10. The answer given here is 50, by the way, but I simply don't know where to start.
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Solving this equation: $3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}$

Solve this equation: $$3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}\qquad (1)$$ I tried to make both sides of the equation have a same base and I started: $$(1)\Leftrightarrow 3^{\log_{4}x}.\sqrt{3}+…
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How to solve this logarithmic system of equations

I was trying to solve this system and I tried to express $y = \frac{100}{x}$ from the first equation and change into the second one and I got $\frac{100}{x}\log_{10}{x} = 10$ After some work I got to $x = 10\log_{10}{x}$ And I cannot solve this…
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Why $\log(x+1)=\log(1+\frac{1}{x})+\log(x)$?

Why is this expression true? $\log(x+1)=\log(1+\frac{1}{x})+\log(x)$ What logarithmic property allows the equivalence?
Kevin
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Solving Problems of the Type $\log(x) = ax^2 + bx + c$

I have recently been struggling to solve the following advanced mathematics problem which is presented as $|0.1 x^2 + 2 x + 3| = \log(x)$. I know that before solving the problem, it must be taken into account that the initial equation must have both…
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Result of square of $(\ln(x))$. [ $(\ln x)^2$ ]

I can’t find the square of $\ln (x)$. How a logarithmic expression be a multiple of itself? $$(\ln(x))^2=?$$
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Logarithm problem

I have an easy problem. I can see the answer but I don't know how to solve the problem and get the answer "the mathematical way". The statement is: $x \large {\cdot 2^{\log _x 5 } = 10 }$ Then I brought it to the form $\log _x 25^x = 10$ The answer…
Andrew
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