Question

16*3^(4x-3) = 3*2^(8x-4)

Answer 1

is this what the question look like
\[16\times3^{(4x-3)}=3\times2^{(8x-4)}\]

Answer 2

yes

Answer 3

ok, do you know the rules on indices?

Answer 4

no

Answer 5

alright, we'll work through it step by step then
first you need to solve 16x3
and 3x2

Answer 6

49 and 6

Answer 7

48

Answer 8

yep so it should now look like
\[48^{(4x-3)}=6^{(8x-4)}\]

Answer 9

now the rules of indices when dealing in powers is that when the base are equal you can now eliminate the base and only work with the powers example
\[3^{z}=3^{2}\]
the because both side has a base number of 3
then z=2

Answer 10

got that?

Answer 11

yes

Answer 12

so now we have to try make the base on both sides equal
that is 48 and 6

Answer 13

so we have to rewrite48 and 6 so that they have the same base?

Answer 14

yeah, something like that. it might be a little had since both numbers cannot be squared

Answer 15

wait a minute lets take a step back actually

Answer 16

ok

Answer 17

see how you have
16x2 and 3x2 at first
we have 2 as base in both

Answer 18

where did you 16x2

Answer 19

get

Answer 20

i mean 16x3. sorry typing error

Answer 21

ok

Answer 22

so if we simplify 16 to \[2^{4}\]
and 3x 2 stay s that way both base becomes
\[3\times2^{4} and 3\times2\]

Answer 23

so it is now
\[3\times2^{4(4x-3)}=3\times2^{(8x-4)}\]

Answer 24

do you get it?

Answer 25

now we can do 4x-3 = 8x-4 ?

Answer 26

yeah but remember the 4 so it should be 4(4x-3)=8x-4

Answer 27

oh ok 16x-12 = 8x-4 so x = 1

Answer 28

is that right?

Answer 29

absolutely..

Answer 30

ok thanks!

Answer 31

so u can now solve for x

Answer 32

your welcome