how to solve this logarithmic equation ?

[atomic zodiac]

Maybe use ln() on both sides and then evaluate the linear equation..

[sharp zodiac]

sorry, but can u show me how?

[atomic zodiac]

Notice:

$\ln(x^a) = aln(x)$

[queen latch]

adonhs

[atomic zodiac]

Well if you use ln() on both sides you get:

$\ln(2^{x-1}) = ln(3^{2x-1})$

[queen latch]

adonhs

[atomic zodiac]

And then you can use the rule from above

was this helpful?

[sharp zodiac]

after that we add (e) to both side to cancel the ln?

[atomic zodiac]

But why would you? You would end up with an exponential equation again

[sharp zodiac]

so what is the next step to solve this

[atomic zodiac]

Well how far are you? Did you end up with some linear equation?

[sharp zodiac]

gimme 1 min i will show u

[atomic zodiac]

your ln(2) became suddenly 2

ln(3) became suddenly 3 how come?

[sharp zodiac]

Bcz i add e to both sides to cancel the ln

[atomic zodiac]

That's not how it works, if you use e properly on both sides you end up with:

$\e^{(x-1)ln(2)} = e^{(2x-1)ln(3)}$

[queen latch]

adonhs

[sharp zodiac]

Ohh

Then how

[atomic zodiac]

well ln(2) and ln(3) are just normal numbers right?

[sharp zodiac]

Right

[atomic zodiac]

the same way you mulitplied with 2 and 3 do the same with ln(2) and ln(3)

[sharp zodiac]

Then i distribute the numbers?

[atomic zodiac]

yea

[sharp zodiac]

Okaaaaayyy

It was that simple omg

[atomic zodiac]

yeah because you end up with a linear equation

the solution may look complicated because ohhhh there is ln but that's ok

[sharp zodiac]

Thank your very much for your time i really appreciate it

[sharp zodiac]

Yea i was confused