Simultaneous Equation involving logs and exponents

[James10492]

9^x-2 = 3^y and log₃2x = 1 + log₃y

Apply 3 to log₃2x = 1 + log₃y to remove the logs:

2x = 3¹ + y

so y = 2x - 3


apply log₃ to 9^x-2 = 3^y, substituting the value for y and you get

log₃9^x-2 = 2x-3

log₃9^x-2 = 2 (x-2)

but now 2x-2 = 2x -3 produces a paradox. What am I doing wrong?

 

[Deleted member 4993]

9^x-2 = 3^y and log₃2x = 1 + log₃y

Apply 3 to log₃2x = 1 + log₃y to remove the logs:

2x = 3¹ + y

so y = 2x - 3

apply log₃ to 9^x-2 = 3^y, substituting the value for y and you get

log₃9^x-2 = 2x-3

log₃9^x-2 = 2 (x-2)

but now 2x-2 = 2x -3 produces a paradox. What am I doing wrong?

You say:

Apply 3 to log₃2x = 1 + log₃y to remove the logs ------ you cannot do that

log₃(2x) = 1 + log₃(y)

log₃(2x) - log₃(y) = 1

log₃(2x/y) = log₃(3)

2x/y = 3

2x = 3y ..................................... now continue

 

[lex]

Apply 3 to [MATH]\log_3(2x)=1+\log_3(y)[/MATH] to remove the logs:

[MATH]2x = 3^1 + y[/MATH]

Yes, that line should be:
[MATH]2x = 3^1 \boldsymbol{\times} y[/MATH]so [MATH]2x=3y[/MATH]

 

[James10492]

Thanks for your replies. I should have tried the method of subtraction using the laws of logs but I was baffled as to why what I was doing did not work.

Yes, that line should be:
[MATH]2x =[B] [/B]3^1 \boldsymbol{\times} y[/MATH]so [MATH]2x=3y[/MATH]

Why do 3 and y become a product?

 

[lex]

Thanks for your replies. I should have tried the method of subtraction using the laws of logs but I was baffled as to why what I was doing did not work.

Why do 3 and y become a product?

[MATH]\log_3{\left(2x\right)}=1+\log_3{\left(y\right)}[/MATH]You do 3 to the power of both sides:
[MATH]3^{\log_3{\left(2x\right)}}=3^{\left(1+\log_3{\left(y\right)}\right)}[/MATH][MATH]2x=3^{(1)}\times3^{\log_3{\left(y\right)}}[/MATH] using the rule of Indices [MATH]a^{m+n}=a^m\times\ a^n[/MATH][MATH]2x=3y[/MATH]

 

[James10492]

thanks Lex for that detailed explanation. It was because I was doing 3^1 + 3^log₃y

 

[lex]

thanks Lex for that detailed explanation. It was because I was doing 3^1 + 3^log3y

Yes. If you write out the step before calculating:
[MATH]3^{\log_3{\left(2x\right)}}=3^{\left(1+\log_3{\left(y\right)}\right)}[/MATH]you're less likely to make a mistake like that.
Hope it's ok now.

 

[Steven G]

9^x-2 = 3^y
(3²)^(x-2)= 3^y

3^2(x-2)=3^y
y = 2(x-2)