Question

Algebra help!! ^_^

Answer 1

i can try

Answer 2

?

Answer 3

i ccccccan try to help

Answer 4

lets do A first.

Answer 5

Oka

Answer 6

okay**

Answer 7

okay so what do you think you do first

Answer 8

simplify the 6^3x

Answer 9

...

Answer 10

\(\bf 6^{3x}=104\qquad \textit{let us take say "log" to both sides}
\\ \quad \\
log(6^{3x})=log(104)\implies \textit{any ideas?}\)

Answer 11

\(\bf recall\implies log(a^{\color{red}{ x}})={\color{red}{ x}}log(a)\)

Answer 12

so.... any ideas on the left-hand-side?

Answer 13

Sorry OS wasn't working for me earlier o-o

Answer 14

3x=log base 6 (104)

Answer 15

I wanna know how to do it though

Answer 16

6^3x = 104
now there is a relation between exponents and logs

Answer 17

idk

Answer 18

b^y= x
log x to the b = y

Answer 19

so no?

Answer 20

What?

Answer 21

no they dont

Answer 22

http://www.purplemath.com/modules/logs.htm
^
The above site might be helpful.

Answer 23

for B
\[\log_{a}x + \log_{a} y = \log_{a} xy \]

Answer 24

rish I think you would know by now that math is not one of my good subjects xDD

I don't understand what you put...

Answer 25

\[\log_{a} x + \log_{b}y = \log_{a}xy\]log to the base a x

Answer 26

When there ia + in log we multiple

Answer 27

is**

Answer 28

@LeilaJudeh ?

Answer 29

@LeilaJudeh u there???????o_O

Answer 30

a)
\[6^{3x}=104\]
Let's take log to both the sides.I'll be using log to the base 10 as you can easily find it's values from a log table given in your text book.
\[\log_{10}6^{3x}=\log_{10}104\]
Rule/Property of log, \[\log_{a}x^n=n \times \log_{a}x\]
Your LHS becomes
\[3x \times \log_{10}6=\log_{10}(13 \times 8)\]
Another rule\[\log_{a}(x \times y)=\log_{a}x+\log_{a}y\]
Apply on RHS\[3x \times \log_{10}6=\log_{10}13+\log_{10}8\]\[3x \times \log_{10}6=\log_{10}13+\log_{10}(2^3)\]
Apply the first rule/property on second term on RHS
\[3x \times \log_{10}6=\log_{10}13+3\log_{10}2\]
\[x=\frac{\log_{10}13+3\log_{10}2}{3\log_{10}6}\]
This way you can calculate by using values for the logs given in log table
You can further simplify denominator
\[x=\frac{\log_{10}13+3\log_{10}2}{3(\log_{10}2+\log_{10}3)}\]
As I remember values for log2 and log3, they are
\[\log_{10}2=0.3010\]\[\log_{10}3=0.4771\]
You can use them but you'll have to look up the value for
\[\log_{10}13\]
Using the same laws and properties attempt your (b) part