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3 years ago

What is the solution oflog6x + log6(x + 5) = 2?

Mathematics

2 answers:

DedPeter [7]3 years ago

7 0

Answer:

The answer is 4 on edg

Step-by-step explanation:

I just answered it :)

exis [7]3 years ago

4 0

Answer:

Doesn't have solution.

Step-by-step explanation:

x ≠ 0; x > 0 (I)

x - 5 > 0

x > 5 (II)

$log_6 x + log_6 (x + 5) = 2 \\\ log_6 [x(x + 5)] = 2 \\\ x^2 + 5x = 36 \\\ x^2 + 5x - 36 = 0 \\\ \Delta = 5^2 - 4.1.(- 36) \\\ \Delta = 169 \\\ \sqrt{\Delta} = \pm 13 \\\ x' = \frac{- 5 + 13}{2} \\\\ x' = 4 \\\ x" = \frac{- 5 - 13}{2} \\\\ x" = - 9$

Then, this expression doesn't have solution.

I Hope I've helped you.

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Basile [38]

Answer:

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Step-by-step explanation:

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4 years ago

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3 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

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The Lagrangian is

$L(x_1,\ldots,x_n,\lambda_1,\ldots,\lambda_n)=x_1+\cdots+x_n+\lambda_1({x_1}^2+\cdots+{x_n}^2)+\cdots+\lambda_n({x_1}^2+\cdots+{x_n}^2)$

with partial derivatives (set equal to 0)

$\dfrac{\partial L}{\partial x_i}=1+2x_i(\lambda_1+\cdots+\lambda_n)=0$

$\dfrac{\partial L}{\partial\lambda_i}={x_1}^2+\cdots+{x_n}^2-36=0$

for each $1\le i\le n$ .

Let $\Lambda$ be the sum of all the multipliers $\lambda_i$ ,

$\Lambda=\displaystyle\sum_{k=1}^n\lambda_k=\lambda_1+\cdots+\lambda_n$

We notice that

$x_i\dfrac{\partial L}{\partial x_i}=x_i+2{x_i}^2\Lambda=0$

so that

$\displaystyle\sum_{i=1}^nx_i\dfrac{\partial L}{\partial x_i}=\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0$

We know that $\sum\limits_{i=1}^n{x_i}^2=36$ , so

$\displaystyle\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0\implies\sum_{i=1}^nx_i=-72\Lambda$

Solving the first $n$ equations for $x_i$ gives

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and in particular

$\displaystyle\sum_{i=1}^nx_i=-\dfrac n{2\Lambda}$

It follows that

$-\dfrac n{2\Lambda}+72\Lambda=0\implies\Lambda^2=\dfrac n{144}\implies\Lambda=\pm\dfrac{\sqrt n}{12}$

which gives us

$x_i=-\dfrac1{2\left(\pm\frac{\sqrt n}{12}\right)}=\pm\dfrac6{\sqrt n}$

That is, we've found two critical points,

$\pm\left(\dfrac6{\sqrt n},\ldots,\dfrac6{\sqrt n}\right)$

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4 years ago

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irakobra [83]

Answer:

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