Solving Logarithmic Equation: Log81x + logx3 = 2

[valon92]

The equation is the following:
Log₈₁x + log_x3 = 2

I've been bashing my head with change of base but i can't seem to get anywhere with it. 81 is rewritable as 3⁴ which seems like it has something to do with the solution, but i just can't seem to get the hang of it. I always get stuck with logs on both sides of the equation and no way to simplify it further..
Any help?
Thanks

 

[Mentallic]

Well the change of base formula is as follows: log_ab=\frac{log_cb}{log_ca} so really you don't even need to use the fact that 81=3⁴, but you can. Use the change of base first and then see where you get.

 

[valon92]

yup using the change of base i got to: ln x/ln 81 + ln3/lnx = 2 but that's as far as i got. I tried simplifying them, getting a common denominator, bringing onto the other side but seems nothing that i do gets me to a solution

 

[Mentallic]

Sorry for the late reply, try multiplying through by log(x) and let log(x)=u

 

[valon92]

Do you mean after the change of base or through the original?

 

[Mentallic]

After the change of base, you'll have a quadratic in u to solve.

And by the way, since log_ab=c is equivalent to a^c=b then log_{a^n}b=c is equivalent to (a^n)^c=b or a^{nc}=b so you can change it into log_ab=nc. You can use this for 81=3^4

 

[valon92]

After the change of base, you'll have a quadratic in u to solve.

And by the way, since log_ab=c is equivalent to a^c=b then log_{a^n}b=c is equivalent to (a^n)^c=b or a^{nc}=b so you can change it into log_ab=nc. You can use this for 81=3^4

solved it using the quadratic equation and replacing ln x with U after the change of base, thanks for the help!

 

[Mentallic]

No problem And I'd suggest you try learn to solve them without using the change of variable, because this helps you notice when you have a quadratic in some function u=f(x). It will definitely come in handy.

 

[valon92]

Any suggestions about expanding
(1+ax)^3/2
for 5 non-zero terms?
using pascal's triangle I am gettin horrible exponents while if i rewrite it as (Root of [1+ax])^3 it doesn't look much better as i don't have two clear values to use in the formula :\

 

[valon92]

uhm, might have solved it: does this look correct?
1+3/2 ax + 3/8 a^2x^2 - 1/16 a^3x^3 + 1/128 a^4x^4

 

[Mentallic]

You should create a new thread so others can look at your question too, because I don't know how to answer it.