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1. Find $x$ if $\log_9(2x-7) = \dfrac{3}{2}$.

2. Calculate $\sqrt{10p} \cdot \sqrt{5p^2} \cdot \sqrt{6p^4}$ . Express your answer in simplest radical form in terms of $p$.  Note: When entering a square root with more than one character, you must use parentheses or brackets. For example, you should enter $\sqrt{14}$ as "sqrt(14)" or "sqrt{14}".

3. Find all solutions $x$ of the inequality $$\frac{5}{24} + \left|x-\frac{11}{48}\right| < \frac{5}{16}.$$Express your answer in interval notation, simplifying all fractions in your answer.

Guest Apr 15, 2018
 #1
avatar+945 
+2

1. log9(2x-7)=1.5

 

logab = c means that a= b

 

So this equation means that:

 

9 to the power of 1.5 is equal to (2x-7)

 

9^1.5 = 2x - 7 

 

Since 9^1.5 = 27, 2x - 7 = 27

 

Solving for x, we have that x = 17.

GYanggg  Apr 15, 2018
 #2
avatar+945 
+2

2. √(10p) * √(5p^2) * √(6p^4)

 

= √(300p^7)

 

= 10p^3√(3p)

GYanggg  Apr 15, 2018
 #3
avatar+945 
+2

3. 

 

1/3 < x < 1/8

GYanggg  Apr 15, 2018
 #4
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0

What is it in interval notation?

Guest Apr 15, 2018
 #5
avatar+90088 
+1

\( \frac{5}{24} + \left|x-\frac{11}{48}\right| < \frac{5}{16}\)

 

subtract 5/24 from both sides

 

l x - 11/48  l  < 5/16 - 5/24

 

l x - 11/48 l  <  15/48 - 10/48

 

l x -  11/48 l  <  5/48        we have two equations, here

 

x - 11/48 < 5/48            x  - 11/48 >  - 5/48

                    

                          add  11/48 to  both sides

 

x < 16/48                   x  > 6/48

 

x < 1/3                x > 1/8

 

x  = ( 1/8 ,  1/3)

 

 

cool cool cool

CPhill  Apr 15, 2018

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