Question

@darkknight dun dun dun~

Answer 1

sorry forgot practically all the properties of log .-.

Answer 2

Btw I got the first one tho lol

Answer 3

2)128

Answer 4

thank you, although I got the first xd .-.

Answer 5

2nd one:
\[3^{x+2}=27^{x-3}\]
\[3^{x+2}=3^{3(x-3)}\] because 3^3 = 27
x+2 = 3(x-3)
solve for x

Answer 6

x+2=3x-9
x=5.5

Answer 7

to check if you did it right you can put the value of x back into the equation

Answer 8

both 3798 then blah blah blah xd

Answer 9

*3787

Answer 10

for the 3rd one (top to bottom not left to right)
you can move the -log base 2 10 to the other side, then exponentiate both sides by 2 to get rid of the log base 2

Answer 11

You will be left with \[2x^2+4x-2=10\]
then solve for x

Answer 12

dont forget to plug in value for x back in to double check

Answer 13

can't factor it <.<

Answer 14

quadratic formula, bring everything to 1 side

Answer 15

._. okok one sec

Answer 16

7.
\[(\frac{ 1 }{ 243 })^x=27\]
3|243

Answer 17

\[\frac{ -4 \pm \sqrt(16+96) }{ 4 } \\ \frac{ -4 \pm \sqrt(112 }{ 4 } \\ \frac{ -4 \pm 4\sqrt(7) }{ 4 } \\ -1 \pm \sqrt(7) \]

Answer 18

log(2x+4)-log(3x)=0
\[\log \frac{ 2x+4 }{ 3x }=0\]
\[\frac{ 2x+4 }{ 3x }=1\]
2x+4=3x
3x-2x=4
x=4

Answer 19

okok got that thx xd

Answer 20

\[2x^2+4x-2=10\]\[2x^2+4x-12=0\]
\[x^2+2x-6=0\]
\[x^2+2x+1=6+1\]
\[x+1=\pm \sqrt{7}\]
\[x=-1 \pm \sqrt{7}\]

Answer 21

i had to get my hands on this haha!
\[ \frac{ \log _{3}(3 x+6)}{\log _{3} 81}=1 \]
\[ \overset{\text{get rid of the denominator}}{==========\Rightarrow} \log _3\left(3x+6\right)=\log _3\left(81\right) \Rightarrow \cancel{\log _3}\left(3x+6\right)=\cancel{\log _3}\left(81\right) \]
\[ 3x+6=81 \]
you solve the rest

Answer 22

11.
\[\frac{ \ln (x+7) }{ \ln (2x-3) }=1\]
\[\ln (x+7)=\ln (2x-3)\]
or x+7=2x-3
x=?

Answer 23

@florisalreadytaken x=25
@surjithayer x=10
that was a lot to read through .-. xddd thanks you guys lmao

Answer 24

and then the last one for #12 13/4?

Answer 25

we left one i think: 12)

\[ \frac{\log (4 x+2)}{\log (15)}=1 \]
\[ \overset{ \text{same for tihis one, get rid of the denominator}}{==================\Rightarrow} \log _{10}\left(4x+2\right)=\log _{10}\left(15\right) \Rightarrow \\ \cancel{\log _{10}}\left(4x+2\right)=\cancel{\log _{10}}\left(15\right) \]
\[ 4x+2=15 \]
\[ 4x=15-2 \]
\[ x=\frac{13}{4} \]
\[ x=?? \]

Answer 26

3.25 xd thx

Answer 27

yeah whatever that is -- is there any left one?

Answer 28

nope lol

thanks surjithayer, flor, and dark xddddd <3

Answer 29

yw