logarithmic help?

Question

Solve the equation

1.) Log(x^2)= (log x)^2

2.) e^2x + 2e^x-15=0

3.)log(base 4) X + Log (base 8) X=1

4.) log (base 3) X - log (base 9) (x+42)=0

Answer 1

1.) Log(x^2)= (log x)^2
2 log x = (log x)(log x)
0 = (log x)(log x) - 2 log x
0 = (log x)(log x - 2)

log x = 0
means x = 1

log x - 2 = 0
log x = 2
x = 10^2 = 100

2.) e^2x + 2e^x-15=0
let a = e^x
now you have:
a^2 + 2a - 15 = 0
(a + 5)(a - 3) = 0
a = 3 or -5

so
e^x = 3
x = ln 3

and
e^x = -5
x = ln (-5)

Answer 2

1) log(x^2) = [log(x)]^2

First off, I'm going to assume your log is base 10, because while the other possibility is log e, log e has its own name (ln).

Use the log property on the left hand side that allows us to move a power inside the log outside of the log.

2 log(x) = [log(x)]^2

Move everything to the right hand side.

0 = [log(x)]^2 - 2log(x)

Factor the right hand side.

0 = [log(x)] [log(x) - 2]

Equate each factor to 0.

log(x) = 0
log(x) - 2 = 10

Convert both equations to exponential form.

x = 10^0
log(x) = 12, x = 10^12

x = 1
x = 1000000000000
Therefore, x = { 1, 1000000000000 }

2) e^(2x) + 2e^x - 15 = 0

This is a quadratic in disguise, because e^(2x) = (e^x)^2.

(e^x)^2 + 2e^x - 15 = 0

This is no harder than solving y^2 + 2y - 15 = 0, which factors as (y + 5)(y - 3) = 0, so our above equation factors as

(e^x + 5)(e^x - 3) = 0

Equate each factor to 0,

e^x + 5 = 0
e^x - 3 = 0

e^x = -5
e^x = 3

The first equation will have no solution (the range of e^x is greater than 0, and -5 is not in the range). The second equation, however, will.

e^x = 3 implies x = ln(3)

3) log[base 4](x) + log[base 8](x) = 1

First off, we _cannot_ combine the logs as per the log identity because we have a difference base. What we need to do is use the change of base formula,
log[base c](a) = log[base b](a) / log[base b](c)
Since 4 and 8 are powers of 2, let's use the change of base formula with log base 2.

log[base 2](x)/log[base 2](4) + log[base 2](x)/log[base 2](8) = 1

But, we can calculate log[base 2](4) and log[base 2](8) directly. log[base 2](4) = 2, and log[base 2](8) = 3, so

log[base 2](x) / 2 + log[base 2](x) / 3 = 1

Multiply everything by 6,

3 log[base 2](x) + 2 log[base 2](x) = 6

Move the powers inside the log.

log[base 2](x^3) + log[base 2](x^2) = 6

Combine the logs.

log[base 2](x^3 * x^2) = 6
log[base 2](x^5) = 6

Convert to exponential form.

x^5 = 2^6

Solve for x by bringing both sides to the (1/5) power.

x = (2^6)^(1/5)
x = 2^(6/5)

4) log[base 3](x) - log[base 9](x + 42) = 0

Like the earlier question, use the change of base formula to change the second log to log[base 3].

log[base 3](x) - log[base 3](x + 42)/log[base 3](9) = 0

log[base 3](x) - log[base 3](x + 42) / 2 = 0

Multiply both sides by 2,

2 log[base 3](x) - log[base 3](x + 42) = 0

Move the 2 inside the log,

log[base 3](x^2) - log[base 3](x + 42) = 0

Use the log identity

log[base 3]( (x^2 / (x + 42) ) = 0

Convert to exponential form.

x^2 / (x + 42) = 1
x^2 = x + 42
x^2 - x - 42 = 0
(x - 7)(x + 6) = 0

This implies x = {7, -6} however we're not finished yet because we have to check for extraneous solutions.

We reject x = -6 because if we plug this into our original equation, log[base 3](x) - log[base 9](x + 42) = 0, we will end up taking the log of a negative number, which is not allowed. Reject x = -6, and our only solution is

x = 7

Answer 3

1) as x >0 log x^2 = 2log x
so (log x)^2 -2log x= 0 log x *(log x-2)=0
log x= 0 x= 1 and log x= 2 x= 10^2 =100
2) call e^x = z
z^2+2z-15=0 z=(( -2+-sqrt(64))/2 z= 3 and z= -5
As the exponential is always positive
z= e^x=3 is the only solution so x=ln3
3) lets work in base 2
log(base4) x= log(base2) x / log base2)4 =1/2 log(base2)x as
log(base 2) 4 =2
also log(base 8) x = log(base 2)x / log(base 2)8
= 1/3 log base 2(x)
Now,in base 2
1/2log x+1/3log x =1
log (x)^5/6 = 1 so x^5/6 = 2 ( if log =1 the number is the base)

so x=2^(6/5)
You can do the last one passing all log to base 3

Answer 4

Sorry I forgot Logs.