Homework Help: Quick maths question

1. Nov 30, 2004

roger

hi guys,

the question says : if 3^x = 9^y-1 show that x=2y-2

I'm not sure how to do this

please can you help me ?

thanx

Roger

2. Nov 30, 2004

Gokul43201

Staff Emeritus
Can you think of a trick that reduces exponents to products ?

3. Nov 30, 2004

roger

do i use log ?

im not too sure

4. Nov 30, 2004

hypermorphism

Logarithms are an excellent way to reduce questions about variable powers to more familiar algebra. :)

5. Nov 30, 2004

roger

Can someone explain further please ?

6. Nov 30, 2004

elote

ln(3^x)=ln(9^y-1) using the properties of logarithms we get:
xln(3) = (y-1)ln(9)
solve for x in terms of y
x = (y-1)ln(9)/ln(3)
I'm sure you can do the rest...

7. Dec 3, 2004

uranium_235

You really do not need to use logs for this question at all. It can be solved in one line. Find the common base of the two bases and manipluate the powers accordingly.

8. Dec 3, 2004

futb0l

Yeah... You can see that

$$3^2 = 9$$

so you dont need to use logs at all :)

9. Dec 4, 2004

Raza

Question:
$$3^x=9^y^-^1$$
Show that $$x=2y-2$$

Solution:
$$3^2=9$$
$$3^x=(3^2)^y^-^1$$
Whenever doing these type of problems,always try to get the same base.

You have to multiply 2 by y-1 so you get 2y-2
$$3^x=3^2^y^-^2$$

Look at the exponents and you get:
$$x=2y-2$$

I might be wrong

10. Dec 4, 2004

Nylex

Raza, your working is correct but you're not meant to post full solutions to problems. It's ok to post hints and to correct their working though (after they've shown it of course).

11. Dec 4, 2004

Raza

I know I shouldn't do that but whenever I need help, I hope someone can finish the problem for me with some explaining along the way. It's not because I just want to copy it off but I don't get it until it's fully done.