Question

4^2x+3 =1
and 2^2x+5-8x

Answer 1

\[\LARGE 4^2x+3=1\]
or
\[\LARGE 4^{2x+3}=1\]
and what do we have to do with the second expression ? :)

Answer 2

the second one and we just have to simplify, but I have no idea how to

Answer 3

if you simplyfy this is the answer

Answer 4

5-4x

Answer 5

for the # 2

Answer 6

thank you!

Answer 7

ok... regarding to the equation
\[\LARGE 4^{2x+3}=1\]
we take logs to both sides...
\[\LARGE \log(4)^{2x+3}=\log (1)\]
\[\LARGE (2x+3)\cdot \log(4)=0\]
now we have two cases
\log(4)=0 ---which is not true.
or to solve
2x+3=0
2x=-3
x=-3/2 Which is true...
because:

\[\LARGE 4^{2x+3}=1\]
\[\LARGE 4^{2\cdot (-3/2)+3}=1\]
\[\LARGE 4^0=1\]
1=1
so
\[\LARGE x=-\frac32 \]

Answer 8

welcome

Answer 9

thank you :)

Answer 10

@reticens your second expression is it like:
\[\LARGE 2^2x+5-8x\]
or
\[\LARGE 2^{2x+5-8x}\]
?

Answer 11

the second one except it is =8x and the 8x is equal for all of the equation

Answer 12

\[\LARGE 2^{2x+5}=8x\]
this?

Answer 13

yes!

Answer 14

then forget what jorge said, he meant another way :) ...
\[\LARGE 2^{2x+5}=8x\]
it's complicated ...let me think :(

Answer 15

alright thank you :)

Answer 16

and yes i know :(

Answer 17

well I have to confess that I'm not capable to solve it.
looks like every value of X we want to take... we'll never get the equation right... :(

Answer 18

its okay! I'll just make a guess haha

Answer 19

I'm affraid you can't.... anyway Good Luck, or hold on. Call someone smarter, maybe they can help you out.

Answer 20

alright thank you!