Question

Help Please. 4^2x+3=1

Answer 1

I need help badly in couple of questions, someone please help!!

Answer 2

is the problem

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


OR

is it

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

Answer 3

yeah that second one

Answer 4

Hint:

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

\[\Large 4^{2x+3}=4^0\]

Answer 5

so we are not using that 1?

Answer 6

we are, just a different form of it

4^0 is the same as 1

Answer 7

ok

Answer 8

the bases are the same (4)

so the exponents must be the same

Answer 9

so

2x + 3 = 0

Answer 10

yup i Got the answer thanks :)

Answer 11

ok great

Answer 12

can you please help me with couple of them too?

Answer 13

sure a few more

Answer 14

ok here is another one

Answer 15

\[10^{-3x}\times10^{x}=\frac{ 1 }{ 10 }\]

Answer 16

\[10^{-3x}\times10^{x}=\frac{ 1 }{ 10 }\]

\[10^{-3x}\times10^{x}=10^{-1}\]

\[10^{-3x+x}=10^{-1}\]


I'll let you finish

Answer 17

ok

Answer 18

what do you get

Answer 19

\[x=\frac{ 1 }{ 2 }\]

Answer 20

I need help with couple more pleaseeee!!

Answer 21

\[64\times16^{-3x}=16^{3x-2}\]

Answer 22

-3x + x = -1

-2x = -1

x = 1/2

good

Answer 23

thanks

Answer 24

\[64\times16^{-3x}=16^{3x-2}\]

\[(4^3)\times(4^2)^{-3x}=(4^2)^{3x-2}\]

\[(4^3)\times4^{2(-3x)}=(4)^{2(3x-2)}\]

\[(4^3)\times4^{-6x}=(4)^{6x-4}\]

\[4^{3+(-6x)}=(4)^{6x-4}\]

\[4^{3-6x}=(4)^{6x-4}\]

Answer 25

I got an answer: \[x=\frac{ 7 }{ 12 }\]

Answer 26

good

Answer 27

\[\frac{ 81^{3n+2} }{ 243^{-n} }=3^{4}\]

Answer 28

oh maybe I can do this one let me try first :)

Answer 29

alright, tell me what you get

Answer 30

I got \[-\frac{ 4 }{ 17 }\]

Answer 31

you nailed it

Answer 32

:)

Answer 33

\[81\times9^{-2b-2}=27\]

Answer 34

tell me what you get

Answer 35

ok

Answer 36

oh I got that one \[-\frac{ 3 }{ 4 }\]

Answer 37

good, that's correct

Answer 38

ok

Answer 39

\[\left(\begin{matrix}1 \\ 6\end{matrix}\right)^{3x+2}\times216^{3x}=\frac{ 1 }{ 216 }\]

Answer 40

and what did you get

Answer 41

I m having trouble starting it but still trying

Answer 42

\[\Large \left(\frac{1}{6}\right)^{3x+2}\times216^{3x}=\frac{ 1 }{ 216 }\]


\[\Large \left(6^{-1}\right)^{3x+2}\times(6^3)^{3x}=6^{-3}\]


\[\Large \left(6\right)^{-(3x+2)}\times(6)^{9x}=6^{-3}\]


\[\Large 6^{-(3x+2)+9x}=6^{-3}\]

Answer 43

oh I only messed up with one sign