Question

Intermediate Algebra question in comments, Please help :)

Answer 1

\[64^{3x}=16^{x+1}\]

Answer 2

Hint: Try to make the base a common number ,Can you guess what is it ?

Answer 3

4

Answer 4

iv'e gotten to \[(3x)\log64=(x+1)\log16 \] using the method i was taught.

Answer 5

im not sure what to do after this though

Answer 6

i have to provide the value of x and an estimate to the fourth decimal place

Answer 7

Great
So by making the base similar , we will get :
\[\Large 4^{3(3x)} = 4^{2(x+1)} \]
Since the base = base ,therefore the power = power .
3(3x)=2(x+1)
9x=2x+1
7x=1
x=1/7

Answer 8

okay i see

Answer 9

k, thx

Answer 10

Back :3

Answer 11

:D

Answer 12

\[\Large (3x)\log64=(x+1)\log16\]
Combine similar terms ,
\[\Large \frac{(3x)}{(x+1)}=\frac{\log16}{\log64}\]
Here is a hint that i always do to prevent confusing ,
Let Log16/Log64 = m
therefore 3x/x+1=m
then \[\large xm+m=3x\]
Now combine again the x
\[\Large xm-3x=m\]
Take the x common ,
\[\Large x(m-3)=m\]
Then
\[\Large \color{Red} X=\frac{m}{m-3}\]

Answer 13

Now Substitute The value of m
and get X :)


P.S:Letting Log16/Log64 = m ,Is not necessary ..I did it only to abbreviate it

Answer 14

You got what i said xD ?

Answer 15

sry, i was away for a sec. im not quite sure i understood what u mean by m

Answer 16

Ignore that part ,Just do the Multiply them by each other .

Answer 17

multiply what by what?

Answer 18

sry im slow

Answer 19

lol

Answer 20

(log16/log64)(x+1)=3x

Answer 21

That ^

Answer 22

i see

Answer 23

so what do i do after that?

Answer 24

combine the x together

Answer 25

and take x common
Then Calulate it

Answer 26

calculate

Answer 27

\[\large 64^{3x}=16^{x+1}\]
\[\large (4)^{3(3x)}=4^{2(x+1)}\]
\[\large 9x=2x+2\]
\[\large 7x=2\]
\[\large x=\frac{2}{7}\]

Answer 28

I already did that !

Answer 29

?, two different answers though.

Answer 30

Why would you use logs when you can just do that.

Answer 31

x=1/7

Answer 32

He is right ,i missed the 2
7x=2

Answer 33

That's wrong. YOu didn't expand/distribute properly.

Answer 34

i need it in the format of log(a)/log(b) i think

Answer 35

*sigh

Answer 36

\[64^{3x}=16^{x+1}\]

Answer 37

\[(4^{3})^{3x}=(4^{2})^{x+1}\]

Answer 38

Then
\[x=\frac{2}{7}\]
\[\log x=\log \frac{2}{7}\]
\[\frac{\log x}{\log \frac{2}{7}}=1\]

Answer 39

\[4^{9x}=4^{2x+2}\]

Answer 40

\[9x=2x+2\]
7x=2
x=2/7

Answer 41

\[\log x=\log \frac{2}{7}\]
\[\log x=\log 2 -\log 7\]
I don't know why you need it in that form and you can't get it into that form if you want x as the subject.

Answer 42

@sgta101