help me solve for with logarithmic equations?

Question

whichever one you can help with is awesome...thnx

log 55= x+4

x^10=356

ln x=5

e^x+2= 1

ln 105= x-9

x^e=100

Answer 1

Here are some hints for you:

1. Just subtract 4 from both sides.
2. Take the tenth root of both sides (do both to the 1/10 power)

3. If lnx = 5 that means e^5 = x
4. I will assume the 2 is part of the exponent, since if not e^x would = -1 which is impossible. Since ln(e^(x+2)) is just x+2, take ln of both sides then move the 2 over.

5. Just add 9 to both sides

6 Do both sides to the 1/e power

Hope that helps without giving it all to you

Answer 2

x = log[55] - 4

x^10 = 356
(x^10)^(1/10) = 356^(1/10)
x = 356^(1/10) = 1.799....... using the Microsoft calculator

lnx = 5
e^ln x = e^5
x = e^5

e^x + 2 = 1
e^x = 1-2 = -1
ln[e^x] = x = ln[-1]

ln[105] = x - 9
x = ln[105] + 9

x^e = 100
(x^e)^(1/e) = x = 100^(1/e)

Answer 3

I could answer all of these, but that would be cheating you of your education. Is this a high school math class or collage? If it is a collage requirement, then " TAKE MORE NOTES AND PAY MORE ATTENTION IN LECTURE!!!" Best wishes for a successful future, and I sincerely mean that .Luck to you!

Answer 4

1)x=log55-4=-2.26
2)10logx = log 356 log x=0.2551 and x = 1.8
3) x=e^5 =148.41
4) impossible e^x>0 (always) so e^x+2 >2 and can´t be 1
5)4.654=x^-9 ln4.654=-9lnx so lnx=-0.171 and x= 0.843
6)elnx=ln100 so lnx= 1.6942 and x=5.4423
Enough.Next time do your homework

Answer 5

log2 [(2x + a million)(x + 4)] = 2 2^( log2 [(2x + a million)(x + 4)] ) = 2^2 2x^2 + 8x + x + 4 = 4 2x^2 + 9x = 0 x(2x + 9) = 0 2x = - 9 x = - 9/2 x = 0 log2 [(-9 + a million)(-9/2 + 4)] = log2 [(-8)(-a million/2)] = log2 [4] = log2 [2^2] = 2log 2 [2] = 2

Answer 6

ln x = 5

The base of ln is e.
You cross the e over to the other side.
It then becomes x= e^5.