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3 years ago

What is the value of x for log6(5x) + log6(2)=5

1 answer:

8 0

$\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log_6(5x)+\log_6(2)=5\implies \log_6(5x\cdot 2)=5\implies \log_6(10x)=5 \\\\\\ 6^{ \log_6(10x)}=6^5\implies 10x=6^5\implies x = \cfrac{6^5}{10}\implies x = \cfrac{3888}{5}\implies x = 777.6$

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This is Matrix for pre calc

gavmur [86]

The given system of equations in augmented matrix form is

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]$

If you need to solve this, first get the matrix in RREF:

Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]$

Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]$

Add 164(row 3) to -91(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]$

Multiply row 4 by 1/13080:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]$

Add -153(row 4) to row 3:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]$

Multiply row 3 by -1/91:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add 6(row 3) and -8(row 4) to row 2:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 2 by 1/5:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

$\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 1 by 1/3:

$\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

So the solution to this system is $(w,x,y,z)=(1,-2,4,-3)$ .

6 0

3 years ago

Ken has 4.66 pounds of walnuts, 2.1 pounds of cashewes, and 8 pounds of peanuts. He mixes them together and divides them equally

weeeeeb [17]

0.82 pounds in each bag

6 0

3 years ago

The total cost in dollars to buy uniforms for the players on a volleyball team can be found using the function c= 34.95u +6.25,

Radda [10]

The domain will be the integer numbers between 8 and 12 using the limits in other words 8 ≤ u ≤ 12

this is considering every player will bought only one uniform for the season

8 0

3 years ago

Read 2 more answers

mike bought a 24 pack of mountain dew for $8.40. write and solve an equation to calculate how much each indivdual can soda cost

vekshin1

Answer:

$0.31

Step-by-step explanation:

You need to take the amount of money each pack costs and divide it by the number of cans to find the unit rate. Hope this helps!!

6 0

2 years ago

Is not b please someone help me

Advocard [28]

√j+√j+14=3√j+10
2√j+14=3√j+10
minus 2√j both sides
14=√j+10
minus 10 both sides
4=√j
squaer both sides
16=j

answer is D, 16

8 0

3 years ago