#1

I needed to solve

(27*3^x)^2 = 27^x*3^1/2

and show how i did it using algebra. so far i've got:

(27^2)*3^(2+x) = 3^(3x+1/2)

6561*3^x=3^(3x+1/2)

What now? Can somebody please walk me through it? I think i need to change both sides to logarithms with a base of 3 but cant figure out how to do it.

Doesn't help that i lost one of my maths books either.

(27*3^x)^2 = 27^x*3^1/2

and show how i did it using algebra. so far i've got:

(27^2)*3^(2+x) = 3^(3x+1/2)

6561*3^x=3^(3x+1/2)

What now? Can somebody please walk me through it? I think i need to change both sides to logarithms with a base of 3 but cant figure out how to do it.

Doesn't help that i lost one of my maths books either.

#2

Factoriseee

#3

Factoriseee

explain?

#4

Theres a maths thread somewhere on here.

#5

CCant you graph it?

On graphing: - x = 3.75

On graphing: - x = 3.75

*Last edited by luv090909 at Oct 27, 2009,*

#6

CCant you graph it?

On graphing: - x = 3.75

yeah but cant see how it could help...

#7

hint:

27= 3^3

this must be freshman algebra

27= 3^3

this must be freshman algebra

*Last edited by rx_eb at Oct 27, 2009,*

#8

well i got it down to 135^x + 9^(x^2) = 3^1/2 - 729

...but i havent done math since calculus in high school so i forget all the little shortcuts and sh1t. good luck with this man lol.

oh and that was 4 years ago btw

...but i havent done math since calculus in high school so i forget all the little shortcuts and sh1t. good luck with this man lol.

oh and that was 4 years ago btw

#9

(27*3^x)^2 = 27^x*3^1/2

(3^3*3^x)^2 = 3^3x*3^1/2

3^(6+2x) = 3 ^ (3x+1/2)

6+2x = 3x+1/2

x = 11/2

(3^3*3^x)^2 = 3^3x*3^1/2

3^(6+2x) = 3 ^ (3x+1/2)

6+2x = 3x+1/2

x = 11/2

#10

hint:

27= 3^3

this must be freshman algebra

Well its level 2 Maths Methods in college here and i'm doing it in high school so meh...

yes i get that 3 cubed is 27 but logarithms still screw with my head.

#11

you don't even need logs here, just basic exponential rules will do

and Seerzy is right

and Seerzy is right

#12

It's not about logarithms at all. As long as you make the base the same, you can take the indexes to be equal, I take it your HSC maths exam is tomorrow? Mine is, my Extension 1 maths is tomorrow morning

#13

why are you adding the powers on the outside of the brackets to the powers on the inside? They should be multiplied shouldn't they? Not sure if its a hint but somethings to the power of 1/2 is the same as the square root of that number

#14

(27*3^x)^2 = 27^x*3^1/2

(3^3*3^x)^2 = 3^3x*3^1/2

3^(6+2x) = 3 ^ (3x+1/2)

6+2x = 3x+1/2

x = 11/2

THANK YOU SO MUCH GOOD SIR!!!

I actually see what i did wrong, and realise how to do it now.

I was thinking to make it a logarithm with a base of 3, but the base no. being 3 makes so much more sense... i was fine with gathering the factors and stuff once i got what i had to do.

once again, thankyou

also, rx_eb, thanks for the hint, if it wasnt like quarter to twelve, and i'd done maths in the last week i may have been able to comprehend it.

ALSO

thanks to everyone who tried to help, even if it was further off than what i got. I now believe in the kindness of mankind.

tl;dr

Thanks.

#15

It's not about logarithms at all. As long as you make the base the same, you can take the indexes to be equal, I take it your HSC maths exam is tomorrow? Mine is, my Extension 1 maths is tomorrow morning

actually i have to do the quality assurance test for my teacher for mathematics methods foundation tomorrow, and i missed like a week of school... and all the revision with it.

there are only 5 of us doing this course too.

why are you adding the powers on the outside of the brackets to the powers on the inside? They should be multiplied shouldn't they? Not sure if its a hint but somethings to the power of 1/2 is the same as the square root of that number

when multiplying powers, you add them. dunno if that makes sense but...

*Last edited by texzephyr at Oct 27, 2009,*