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djyliett [7]
2 years ago
15

If log_10(x) = 3 + log_10(y), then find x/y

Mathematics
2 answers:
s344n2d4d5 [400]2 years ago
8 0

Answer:

  x/y = 10^3 = 1000

Step-by-step explanation:

The rules of logarithms tell you ...

  log(x/y) = log(x) -log(y)

__

Here, we have ...

  log(x/y) = log(x) -log(y)

  = (3 +log(y)) -log(y) = 3 . . . . . . . . substittute for log(x)

Taking the antilog, we get ...

  log(x/y) = 3

  x/y = 10^3 = 1000

_____

<em>Additional comment</em>

The expression log(x) is often used to refer to the "common log" of x, which is the logarithm to the base 10. This lets us avoid the cumbersome notation log_10(x).

In more formal mathematics, log(x) may be used to refer to the natural log. The distinction is usually made in high school algebra by referring to the natural log using ln(x).

Lina20 [59]2 years ago
7 0

Answer:

  •  1000

Step-by-step explanation:

  • log_{10}x = 3 + log_{10}y
  • log_{10}x - log_{10}y = 3
  • log_{10}(x/y) = 3
  • x/y = 10^3
  • x/y = 1000

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Answer:

His gain percent is 20%

Step-by-step explanation:

In this question, we want to find the percentage gain of the business.

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Using the values we have in the question;

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I can't figure out how to do (i + j) x (i x j)for vector calc
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In three dimensions, the cross product of two vectors is defined as shown below

\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}

Then, solving the determinant

\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}

In our case,

\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}

Where we used the formula for AxB to calculate ixj.

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Thus, (i+j)x(ixj)=i-j

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