logarithmic equation

[PixiePink]

I need help solving this logarithmic equation
log3(5x+5) - log3(x^2 - 1) = 0

 

[Guest]

log3(5x+5) - log3(x^2 - 1) = 0

log a - log b = log (a/b) when the same base is used. Do this with your equation.

Also (5x+5) = 5(x+1)

and (x^2-1) =(x+1)(x-1)

this allows some cancell out to occur.

Then remove the log and change to power form..
Have a go and I will assist.

 

[PixiePink]

so would it become
5/x-1=3^0
5/x-1=1
5=1(x-1)
5=x-1
6=x

 

[Guest]

well done

 

[soroban]

Hello, PixiePink!

log₃(5x+5) - log₃(x² - 1) = 0
.

I avoid creating quotients (fractions) whenever possible

We have: .log₃(5x + 5) .= .log₃(x² - 1)

Take antilogs: .5x + 5 .= .x² - 1

We have a quadratic: .x² - 5x - 6 .= .0

. . which factors: .(x - 6)(x - 1) .= .0

. . and has roots: .x .= .6, -1

We find that x = -1 is an extraneous root.

. . Therefore, the solution is: . x = 6