Question

Logarithm Help needed, writing question below :) Just need someone to double-check an answer

Answer 1

@KevinOrr

Answer 2

no. you can use a calculator to show it does not work

Answer 3

K, the other answer choices r:

Answer 4

I think I'm doing a mistake*

Answer 5

take the log (base 10 or base e ) of both sides

log ( 3^(1-2x)) = log(2)
(1-2x) log(3) = log(2)
1-2x = log(2)/log(3)
1- log(2)/log(3) = 2x

(1-log(2)/log(3)) /2 = x

Answer 6

$$\bf 3^{1-2x} = 2\\
log_3(3^{1-2x}) = log_32\\
1-2x = \cfrac{log(2)}{log(3)}\\
\cfrac{
1+\frac{log(2)}{log(3)}}{
2} = x
$$

Answer 7

I skipped some steps on the algebra... I assume you follow ?

Answer 8

No wonder I did [1- log(2)/log(3)]/2 for some reason Thx :) gonna figure it out now

Answer 9

hmmm, wait a minute, you're right... should be negative :(

Answer 10

and i flipped the logs around,

Answer 11

$$\bf
3^{1-2x} = 2\\
log_3(3^{1-2x}) = log_32\\
1-2x = \cfrac{log(2)}{log(3)}\\
\cfrac{
1-\frac{log(2)}{log(3)}}{
2} = x
$$

Answer 12

that looks good. [1- log(2)/log(3)]/2

is your calculator broken?

Answer 13

hehe

Answer 14

lol it's hard when u have 2 use the internet as ur calculator and i got x~.185

Answer 15

yes. 0.185

Answer 16

Thx a ton to both of u! :)