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3 years ago

Expand the logarithm. log4 (3xyz)^2

Mathematics

2 answers:

Paladinen [302]3 years ago

8 0

Answer:

Answer is (2log2 + 2log3 + 2logx + 2logy + 2logz)

Step-by-step explanation:

log4(3xyz)²

= log4 + log(3xyz)²

= log4 + 2log(3xyz)

= log(2)² + 2[log3 + logx + logy + logz]

= 2log2 + 2log3 + 2logx + 2logy + 2logz

tangare [24]3 years ago

4 0

2(log4(3)+log4(x)+log4(y)+log4(z))

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|x| + 6 = 3 A question on homework

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Answer:

-9

Step-by-step explanation:

-9+6=-3

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3 years ago

Which inequality has (–2.5, –4.5) as part of the solution set?

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2 years ago

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Explain why x squared = 16 has two solutions. What are the solutions.

BabaBlast [244]

A negative number * another negative number gives a positive value

A positive number * another positive number also gives a positive value

The solutions are -4, 4

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3 0

2 years ago

Help with quadratic graphs asap!

zlopas [31]

x= -0.25 and 1.25

x = -0.5 and 1.5

x = 0 and 1

x = -0.375 and 1.625

Step-by-step explanation:

Given equation of the curve is

$y = 4x^2 -4x-1$

(i)

First equation $4x^2 -4x-1 =0$

Here y = 0

In x-axis the value of y = 0

The graph cuts the x-axis at(-0.25,0) and (1.25,0)

So, x= -0.25 and 1.25

(ii)

$4x^2 -4x-1 =2$

here y = 2

The line y= 2 cuts the curve at (-0.5,2) and (1.5,2)

So, x = -0.5 and 1.5

Given equation of the curve is

$y= 3x^2-3x -1$

(i)

$3x^2-3x +2 =2$

$\Leftrightarrow 3x^2-3x -1 = -1$

Here y = -1

The line y = -1 cuts the curve at (0,-1) and (1,-1)

So x = 0 and 1

(ii)

$3x^2-3x -1 =x+1$

Here y = x+1

The line y -x =1 cuts the curve at (-0.375,0.5) and (1.625,2).

So x = -0.375 and 1.625

3 0

3 years ago

Please someone help me to prove this..

Pachacha [2.7K]

Answer: see proof below

Step-by-step explanation:

Use the following Sum to Product Identities:

$\sin x+\sin y=2\sin \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\sin x-\sin y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=-2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)$

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \qquad \dfrac{\sin 5-\sin 15+\sin 25 - \sin 35}{\cos 5-\cos 15- \cos 25 + \cos 35}$

$\text{Reqroup:}\qquad \qquad \qquad \dfrac{(\sin 25+\sin 5)-(\sin 35 + \sin 15)}{(\cos 35+\cos 5)-(\cos 25 + \cos 15)}$

$\text{Sum to Product:}\quad \dfrac{2\sin \bigg(\dfrac{25+5}{2}\bigg)\cos \bigg(\dfrac{25-5}{2}\bigg)-2\sin \bigg(\dfrac{35+15}{2}\bigg)\cos \bigg(\dfrac{35-15}{2}\bigg)}{2\cos \bigg(\dfrac{25+15}{2}\bigg)\cos \bigg(\dfrac{25-15}{2}\bigg)-2\cos \bigg(\dfrac{35+5}{2}\bigg)\cos \bigg(\dfrac{35-5}{2}\bigg)}$ $\text{Simplify:}\qquad \qquad \dfrac{2\sin 15\cos 10-2\sin 25\cos 10}{2\cos 20\cos 15-2\cos 20\cos 5}$

$\text{Factor:}\qquad \qquad \dfrac{2\cos 10(\sin 15-\sin 25)}{2\cos 20(\cos 15-\cos 5)}$

$\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}$

$\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10$

LHS = RHS: cot 10 = cot 10 $\checkmark$

8 0

3 years ago

Expand the logarithm. log4 (3xyz​)^2

Expand the logarithm. log4 (3xyz)^2