Two logarathmic equations: logbase7of(x) = logbase2of(x-3), and...

[polk416]

Hello, I got two log equations that I am somewhat stuck on.

A) logbase7of(x) = logbase2of(x-3)

or written with the change of base: (logx/log7) = log(x-3)/log2

I completely do not remember how to solve a log equation with different bases.

I tried (lnx/ln7)-(ln(x-3))/ln2). From there it's just a complex fraction, but my answer when plugged back in is incorrect.

B) (5^x)+10*(5^-x)=4

This is what I got

It turns into a quadratic: 5^(2x)-4(5^x)+10=0

Using the quadratic formula it simplifies into 5^x=2+i√6
Take the the natural log of both sides to get x=(ln(2+i√6)/(ln5)
No real solution.



Edit: Added my work

 

[Cartesius24]

Task

1. I see first equation is solvable olny in numeric way.
Let us put\(\displaystyle x= 7^t \)
If we use it We'll make a form
\(\displaystyle t= \log _{2} (7^t - 3) \)
so
\(\displaystyle 2^t = 7^t - 3) \)
2. Yes . I confirm it.

 

[polk416]

Okay, I got B figured out but A keeps going nowhere.

I need it in exact form, and I keep getting to an equation where it is not solvable in exact form.