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Lerok [7]
4 years ago
8

Solve 2^x=3. round to the nearest ten thousandth

Mathematics
2 answers:
nikitadnepr [17]4 years ago
8 0
2^{x}  = 3
log_{2}3 = x
\frac{log_3}{log_2} = x
x = 1.585
Soloha48 [4]4 years ago
5 0

Answer:

The solution of the expression is x=1.584.          

Step-by-step explanation:

Given : Expression 2^x=3

To find : Solve the expression round to the nearest ten thousandth?

Solution :  

Step 1 - Write the expression,

2^x=3

Step 2 - Taking log both side,

\log (2^x)=\log (3)

Step 3 - Apply logarithmic property, \log a^x=x\log a

x\log (2)=\log (3)

Step 4 - Solve,

x=\frac{\log (3)}{\log (2)}

x=\frac{0.477}{0.301}

x=1.584

Therefore, The solution of the expression is x=1.584.

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Tadashi had a veggie pizza that was cut into 8 equal pieces. He ate 2 pieces and left the other 6 in the box. What fraction of t
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A square has a perimeter of 64cm. What is the length of each side? please explain!!!
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4 years ago
What is the best approximate solution to the system of linear equations?
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3 years ago
Two boats leave port at noon. Boat 1 sails due east at 12 knots. Boat 2 sails due south at 8 knots. At 2 pm the wind diminishes
Ivan

Answer:

14.86 knots.

Step-by-step explanation:

<em>Given that:</em>

The boats leave the port at noon.

Speed of boat 1 = 12 knots due east

Speed of boat 2 = 8 knots due south

At 2 pm:

Distance traveled by boat 1 = 24 units due east

Distance traveled by boat 2 = 16 units due south

Now, speed of boat 1 changes to 9 knots:

At 3 pm:

Distance traveled by boat 1 = 24 + 9= 33 units due east

Distance traveled by boat 2 = 16+8 = 24 units due south

Now, speed of boat 1 changes to 8+7 = 15 knots

At 5 pm:

Distance traveled by boat 1 = 33 + 2\times 9= 51 units due east

Distance traveled by boat 2 = 24 + 2 \times 15 = 54 units due south

Now, the situation of distance traveled can be seen by the attached right angled \triangle AOB.

O is the port and A is the location of boat 1

B is the location of boat 2.

Using pythagorean theorem:

\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AB^{2} = OA^{2} + OB^{2}\\\Rightarrow AB^{2} = 51^{2} + 54^{2}\\\Rightarrow AB^{2} = 2601+ 2916 = 5517\\\Rightarrow AB = 74.28\ units

so, the total distance between the two boats is 74.28 units.

Change in distance per hour = \dfrac{Total\ distance}{Total\ time}

\Rightarrow \dfrac{74.28}{5} = 14.86\ knots

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3 years ago
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