1. ## Indices

Need some guidance!!!!!!!!

1)Simplify the following leaving the answer in index form:

a.(53)2

b.(m3)3

c.(a3b3)2

2)Determine the value of:

a)3-3

b)2-4

c)27-1/3

d)125-2/3

3)Solve the following:

a)23x = 128

b)32x = 243

c)54x = 625

4)a) 82x = 1

64

b) 252x = 1

125

c) 493x = 1

343

2. Hi soul

how about you post what you have tried to solve your questions

3. Originally Posted by Soul
Need some guidance!!!!!!!!

1)Simplify the following leaving the answer in index form:

a.$\displaystyle (53)^2$

b.$\displaystyle (m^3)^3$

c.$\displaystyle (a^3b^3)^2$

2)Determine the value of:

a)$\displaystyle 3^{-3}$

b)$\displaystyle 2^{-4}$

c)$\displaystyle 27^{-1/3}$

d)$\displaystyle 125^{-2/3}$

3)Solve the following:

a)$\displaystyle 2^{3x} = 128$

b)$\displaystyle 3^{2x} = 243$

c)$\displaystyle 5^{4x} = 625$

4)a) $\displaystyle 8^{2x} = \frac{1}{64}$

b) $\displaystyle 25^{2x} = \frac{1}{125}$

c) $\displaystyle 49^{3x} = \frac{1}{343}$

Hi Soul,

I've tried to clean up your post a little bit, but I'm not so sure about #3 and #4. Can you help us out a little here? And did I correctly interpret the first 2?

4. and at first i think that question 2a) 3-3 = 0...
silly me

5. You are right, but the number changes when I post them, for question 3
a)23x = 128
b)32x = 243
c)54x = 625

4)a) 82x = 1
64
b) 252x = 1
125
c) 493x = 1
343

6. Originally Posted by Soul
You are right, but the number changes when I post them, for question 3
a)23x = 128
b)32x = 243
c)54x = 625

4)a) 82x = 1
64
b) 252x = 1
125
c) 493x = 1
343
Hi Soul,

If you don't know the code to convert to mathematical symbols, you will have to resort to parentheses and ^ to indicate raising a value to a power. Like....

(2)^(3x)=128

This is interpreted as $\displaystyle 2^{3x}=128$

(25)^(2x) = 1/25

This is interpreted as $\displaystyle 25^{2x}=\frac{1}{25}$

Now, I'll work a few of these and you can try the rest and let us know how well you did.

1a. Are you sure this one is correct? Seems trivial when compared to the others.

$\displaystyle 53^2=2809$

1b. $\displaystyle (m^3)^3=m^{3 \cdot 3}=m^9$

2a. $\displaystyle (3)^{-3}=\frac{1}{(3)^3}=\frac{1}{27}$

2d. $\displaystyle (125)^{-2/3}=(5^3)^{-2/3}=(5)^{-2}=\frac{1}{(5)^2}=\frac{1}{25}$

3a. $\displaystyle 2^{3x}=128$

$\displaystyle 2^{3x}=2^7$

$\displaystyle 3x=7$

$\displaystyle x=\frac{7}{3}$

4a. $\displaystyle 8^{2x}=\frac{1}{64}$

$\displaystyle 8^{2x}=\frac{1}{8^2}$

$\displaystyle 8^{2x}=8^{-2}$

$\displaystyle 2x=-2$

$\displaystyle x=-1$

You should now be able to use these worked examples to help you with the others. Good luck.

7. Originally Posted by masters

$\displaystyle 53^2=2809$
i think it should be still $\displaystyle 53^2$ because the question asks in index form ?

CMIIW

8. Hey ,

Thank you, for the tip on posting my powers and fractions, I will try using it.

9. Originally Posted by songoku
i think it should be still $\displaystyle 53^2$ because the question asks in index form ?

CMIIW
You may be right, songoku, but I think maybe it's....

$\displaystyle (5^3)^2=5^6$

We'll have to watch and wait for the OP to tell us, won't we?

10. Originally Posted by masters
$\displaystyle (5^3)^2=5^6$
Yes it makes sense and i guess this is the right question

btw, what does OP stands for?

11. Originally Posted by songoku
Yes it makes sense and i guess this is the right question

btw, what does OP stands for?
Original Poster!

12. Originally Posted by Soul
Hey ,

Thank you, for the tip on posting my powers and fractions, I will try using it.
And, Soul, if you're planning to post on the forum a lot you will probably want to learn the LaTex code. Start here..

http://www.mathhelpforum.com/math-help/latex-help/

13. This is the right format, I just asked my tutor. (5^3)^2

14. ## Log

1) a) 641/2 x 271/3 x 160 =

b) Compute:

log232 – log3243 =

log464

c) log25k = ½

K =

2) (x3/2)3 x √(x9)

15. Originally Posted by Soul
1) a) 641/2 x 271/3 x 160 =

b) Compute:

log232 – log3243 =

log464

c) log25k = ½

K =

2) (x3/2)3 x √(x9)

DO u mean something else for (a) , or else you can just get the answer from the calculator .

For (b) , i hope u mean :

$\displaystyle log_2{32}-log_3{243}= log_2{2^5}-log_3{3^5}$

Can u simplify it using the log properties ? I'm sure u can .

(c) $\displaystyle log_25{k}=\frac{1}{2}$

Convert this to index form : $\displaystyle 25^{\frac{1}{2}}=k$
$\displaystyle (x^{\frac{3}{2}})^3\times\sqrt{x^9}$ ??

Well i guess so .

$\displaystyle x^{\frac{9}{2}} \cdot x^{\frac{9}{2}}$

do you see why ?

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