ln&log related problems?

Question

solve for x
1.8^x=(1/32)^(x-2)

2. log10 X=-1

I got x=.1 correct?

3.(1/3)e^(-3/x)=0.9

Use the laws of logarithms to simplify the given expressions
1.lnx e^(-1/2)

2. lnx(2x-1)(x-1)

3. log ([squ(x+1)]/(x^2+1))

4. log x (x^2+1)^(-.5)

5. ln(2x-1)(x+1)



Condense the following

1.2[ln x-ln(x+2)-ln(x^2-3)]

2. 2[ln 8- ln(x^2+1)]

3. ln3+1/3ln(4 - x^2) - ln x


Thanks!

Answer 1

While agreeing with Helmut, I'll give a bit of help. After all, you did get question 2 correct!

1. Notice 8 and 32 are both powers of 2. So you can write this as
(2^3)^x = (2^(-5))^(x-2)
In this case we multiply the exponents together, giving
2^(3x) = 2^(-5x+10)

If two powers of 2 are equal, the exponents must be equal, so if you solve
3x = -5x + 10 you have the answer.

Qu.3:
First multiply both sides by 3, then use the fact that
if e^x = y, then
x = ln y.

Thus you find what -3/x is equal to. To solve for x, multiply both sides by x and divide both sides by ln 2.7. Not a very attractive answer, but it's right.

I'm a bit puzzled about the next set. What is the x for in
lnx e^(-1/2)?
Is this the natural logarithm of (x multiplied by e^(-1/2))? or is it just e^(-1/2) multiplied by the natural logarithm of x? Or should the x not be there?

Note that ln e^(-1/2) = -1/2, because
ln is the inverse function of e^, so they cancel each other out in the sense that
ln (e^a) = a, and also e^(ln a) = a.
It's just like (√a)^2 = a, because square and square root are inverses of each other; and x + a - a = x because adding and subtracting are inverses of each other.

If that first one is
ln(x e^(-1/2)), then you can write it as
ln x + ln (e^(-1/2)
= ln x - 1/2

For the second one use the rule
log ab = log a + log b
to write it as
ln x + ln (2x - 1) + ln (x - 1), though I wouldn't call this any simpler than the given expression.

For the third, you are probably expected to recognise that the square root is represented by exponent 1/2, so write

log ([√(x+1)]/(x^2+1))
= log [(x + 1)^(1/2)]/(x^2 + 1)
then use the rule
log a/b = log a - log b and also
log p^(1/2) = (1/2) log p.

If this is helping, and you're getting the hang of them now, you should be able to do some more for yourself.

Answer 2

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Answer 3

I already did homework like this years ago. I see no reason to do yours and let you present it to your teacher as your work.