Question

can someone please explain

3^(2x+1)=2(5^x) ?? thank you!!!

Answer 1

Is this your question?
\[3^{(2x+1)}=2\times 5^x\]

Answer 2

\[3^{2x+1}=2(5^x)\]

Answer 3

yup!

Answer 4

Do you know logarithms?

Answer 5

@dydlf ?

Answer 6

yup! that's our lesson too!:)

Answer 7

Ok, Let's take logarithm to the base 10 both sides
\[\log(3^{2x+1})=\log(2(5^x))\]
Now we'll use two important properties of logarithms to simplify this
\[\log (A\times B)=\log A+\log B\]
\[\log (A^a)=a\times \log A\]
Do you know these properties?

Answer 8

yup i know those properties! :D wait i'll be back

Answer 9

@dydlf could you try to simplify using these properties?

Answer 10

\[\frac{ (3^{2x+1})}{ 5^x }=2\] then i multiplied the left with log base 5?!?! is that legal if ever?

anyway i tried this too... i divided 2 in both sides soooo \[\frac{ 3 }{ 2 } ^{2x+1}=5^x\]
theeen... i did log with base 5 soooo i ended with x at the other side then i brought the exponent 2x+1 to the left so i ended up with log 3/2 base 5=x/2x+1 ?? :( and i can't solve for x anymore :(

Answer 11

Multiplying log won't help, I'll take log ( base 10) both sides
\[\log(3^{2x+1})=\log(2(5^x))\]
Now we'll use the properties to simplify this
\[(2x+1)\times \log 3=\log 2+\log 5^{x}\]

Answer 12

Do you get this?
I have used these properties
\[\log AB=\log A+\log B\]
\[\log A^a=a\log A\]

Answer 13

OOOOHHH i'm seeing something wait

Answer 14

ok

Answer 15

is it x= log2-log3/2log3-log5 ??

Answer 16

Yes, you're right

Answer 17

yay thank you!!!