Solve the equation- help (1 Viewer)

[Mohit7]

How do i solve this equation:

2lnx=ln(5 + 4x)

Thanks

 

[Captain Gh3y]

take e to the power of both sides

e^(2ln(x)) = e^(ln(5+4x))

x^2 = 5+4x

Solve quadratic from here, ignore solutions not greater than zero

 

[watatank]

2lnx=ln(5 + 4x)

2=ln(5 + 4x)/lnx

log_x(5 + 4x) = 2

x²= (5 + 4x)

x²- 4x - 5 = 0

(x-5)(x+1) = 0

x = -1,5

but when x = -1, there is no solution.

x = 5

 

[watatank]

take e to the power of both sides

e^(2ln(x)) = e^(ln(5+4x))

x^2 = 5+4x

Solve quadratic from here, ignore solutions not greater than zero

fuck i knew there was an easier way. do it that way, my way is the way of suckers

 

[eam]

take e to the power of both sides

e^(2ln(x)) = e^(ln(5+4x))

x^2 = 5+4x

Solve quadratic from here, ignore solutions not greater than zero

Im confused, how and why do u take e to the power of both sides?

 

[Captain Gh3y]

Because e^(ln(x)) = x ?

 

[MAlame]

thats right!!!

 

[Slidey]

You can take the ln of something or raise it to base e, as many time as you want, as long as you do it to both sides... unless you do both at once, in which case it doesn't matter if you only do it to one side, as these operations cancel each other out.

 

[Forbidden.]

To prove e^{ln x} = x

y = e^{ln x}

ln y = ln e^{ln x}

ln y = ln x . ln e (Logarithm Law log_axⁿ = n log_ax)

ln y = ln x (Because ln e = 1)

y = x (ln's cancel out)

e^{ln x} = x (y = e^{ln x})

 

[Martyno1]

To prove e^{ln x} = x

y = e^{ln x}

ln y = ln e^{ln x}

ln y = ln x . ln e (Logarithm Law log_axⁿ = n log_ax)

ln y = ln x (Because ln e = 1)

y = x (ln's cancel out)

e^{ln x} = x (y = e^{ln x})

ugh why did I not know that? :|
Thanks, that helped more than 1 person.