Solving 3^x = 9^y-1: Easy Math Question Explained by Roger

[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

 

[Gokul43201]

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

 

[roger]

do i use log ?

im not too sure

 

[hypermorphism]

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

 

[roger]

Can someone explain further please ?

 

[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...

 

[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.

 

[futb0l]

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.

Yeah... You can see that

3^2 = 9

so you don't need to use logs at all :)

 

[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

 

[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).

 

[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.