Thread: solving for x

okay doing logarithms and exponentials, i have to solve for x
im having trouble with 2 of these problems

1. e^4x - e^2x - 30 = 0
i know i have to use the quadratic formula for this one and it was going well until i tried to factor them all out. i got to here: (e^2x)^2 - e^2x - 30 = 0

how do i factor them out?

2. log(4x) - log(12+ square root of x) = 2
to me that just looked flat out intimidating =/

Originally Posted by ohchelsea

okay doing logarithms and exponentials, i have to solve for x
im having trouble with 2 of these problems

1. e^4x - e^2x - 30 = 0
i know i have to use the quadratic formula for this one and it was going well until i tried to factor them all out. i got to here: (e^2x)^2 - e^2x - 30 = 0

how do i factor them out?

ok, apparently seeing the e^2x is making this problem seem too complicated to you. let's replace e^2x with y and see if you follow.

let e^2x be y, then we have

y^2 - y - 30 = 0 .............does it look simplier now? it should, this is a regular (and simple) quadratic equation.

=> (y - 6)(y + 5) = 0
=> y = 6 or y = -5

but y is e^2x
=> e^2x = 6 or e^2x = -5

e^2x can never be -5 so there is no solution for that equaition, so
=> e^2x = 6 ............this is an exponential equaiton, we proceed by logging both sides
=> lne^2x = ln6
=> 2x = ln6
=> x = (ln6)/2

2. log(4x) - log(12+ square root of x) = 2
to me that just looked flat out intimidating =/

this is log to the base 10 i assume

log(4x) - log(12 + sqrt(x)) = 2 .........let's combine the logs on the left
=> log[(4x)/(12 + sqrt(x))] = 2 ........since logx - logy = log(x/y)
=> 10^2 = (4x)/(12 + sqrt(x)) .........since if log[a] b = c, then a^c = b (log[a]b means log to the base a of b)
=> 100 = 4x/(12 + sqrt(x))
=> 120 + 100sqrt(x) = 4x
=> 4x - 100sqrt(x) - 120 = 0 .........that itself is complicated! Now what?!!

calm down, let's rewrite this

4x - 50*2sqrt(x) - 120 = 0 ..........are you seeing it yet? it's similar to the first one

=> (2sqrt(x))^2 - 50(2sqrt(x)) - 120 = 0 .........see it now?

let 2sqrt(x) be y
=> y^2 - 50y - 120 = 0
for some reason i'm having trouble seeing the factors for this, let's use the quadratic formula

=> y = [50 +/- sqrt(2500 + 480)]/2
=> y = [50 +/- sqrt(2980)]/2 ...........no wonder i could not see the factors, there's no nice number for that!

but y = 2sqrt(x)
=> 2sqrt(x) = [50 +/- sqrt(2980)]/2

but 2sqrt(x) can't be negative, so we only want the positive answer, that is,

2sqrt(x) = [50 + sqrt(2980)]/2
=> sqrt(x) = 50 + sqrt(2980)
=> x = [50 + sqrt(2980)]^2

so just work that out

okay for the second one would 120 be 1200 instead since you're multiplying 12 by 100?

Originally Posted by ohchelsea

okay for the second one would 120 be 1200 instead since you're multiplying 12 by 100?

you are correct sir! Congrats on catching that! it shows you were not just mindlessly copying down the answer. make the necessary changes.

Hello, ohchelsea!

1. .e^4x - e^2x - 30 .= .0

i know i have to use the quadratic formula for this one and it was going well
until i tried to factor them all out.

i got to here: .(e^2x)² - e^2x - 30 .= .0

how do i factor them out?

Replace e^2x with u.

We have: .u² - u - 30 .= .0

. . which factors: .(u + 5)(u - 6) .= .0

. . and has roots: .u .= .-5, 6


We have: .e^2x = -5 . . . which has no real roots.

. . . And: .e^2x = 6 . → . 2x = ln(6) . → . x = ½·ln(6)

2. log(4x) - log(12+ √x) .= .2

We have: .ln(4x) .= .2 + log(12 + √x) .= .log(100) + log(12 + √x) .= .log[100(12 + √x)]

. . 4x .= .100(12 + √x) .= .4x .= .1200 + 100√x . . → . . x - 300 .= .25√x

Square both sides: .(x - 300)² .= .(25√x)²

. . x² - 600x + 90,000 .= .625x . . → . . x² - 1225x + 90,000 .= .0
. . . . . . . . . . . . . . . . . . . . . . . . . . . __
. . . . . . . . . . . . . . . . . . 1225 ± 125√73
Quadratic Formula: . x. = .----------------- . ≈ . 78.5, 1146.5
. . . . . . . . . . . . . . . . . . . . . . .2


The first root is extraneous. .Therefore: .x .≈ .1146.5

Originally Posted by Jhevon

...
2sqrt(x) = [50 + sqrt(2980)]/2
=> sqrt(x) = 50 + sqrt(2980)
...

Hello, Jhevon,

it's me again and I've a problem with your transformation.

I assume that you divided both sides of the equation by 2. Then the result should be:

=> sqrt(x) = (50 + sqrt(2980))/4

Originally Posted by earboth

Hello, Jhevon,

it's me again and I've a problem with your transformation.

I assume that you divided both sides of the equation by 2. Then the result should be:

=> sqrt(x) = (50 + sqrt(2980))/4

you are correct earboth. i hope ohchelsea caught that