logarithms! ... help!?

Question

1. 3^1-2x = 4^x

2. (3/5)^x = 7^1-x

3. 1.2^x = (0.5)^-x

4. 5(2^3x) = 8

5. log(base a) (x-1) - log(base a) (x+6) = log(base a) (x-2) - log(base a) (x+3)

6. log(base 1/3) (x^2 + x) - log(base 1/3) (x^2 - x) =-1

No, I do not want you to do my homework .. I am trying to study for my test, and stuck on these problems! (answering c...)

^i was answering zerocarbon...

Answer 1

3^(1-2x) = 4^x
(1-2x)ln3 = xln4
x = ln3 / (ln4+2ln3)

[I assume you intended brackets round the 1-2x as equations with a^x and x^r terms are particularly difficult to solve except numerically ... although in this case I can spot that x=1/2 would be a solution of (3) - 2(1/2) = 4^(1/2) ]


(3/5)^x = 7^(1-x)
x ln(3/5) = (1-x) ln7
x = ln7 / (ln7+ln(3/5) )


1.2^x = 0.5^(-x)
x ln1.2 = -x ln0.5
x=0


5(2^(3x)) = 8
ln5 + (3x)ln2 = ln 8
x = (ln8-ln5)/(3ln2)


log[a](x-1) - log[a](x+6) = log[a](x-2) - log[a](x+3)
log[a](x-1) + log[a](x+3) = log[a](x-2) + log[a](x+6)
(x-1)(x+3) = (x-2)(x+6)
x^2 + 2x -3 = x^2 + 4x -12
2x = 9
x = 4.5


log[1/3](x^2+x) - log[1/3](x^2-x) = -1
(x^2+x) / (x^2-x) = 1/(1/3) = 3
2x^2-4x =0
so x=0 or x=2
but x=0 disallowed by original equation (log0)
so x = 2

Answer 2

If 3^(1-2x) = 4^x
(1-2x)Ln(3) = xLn(4),
giving you a linear equation in one variable. The rest are pretty much the same.
In #5 once you combine terms, the log(base a) can be dropped. (Log(x) - Log(y) = Log(x/y) holds for any base.

Answer 3

5 × 2^3x = 8 (divide both sides by 5)
2^3x = 1.6 (take the log(base10) of both sides)
log(base10)2^3x) = log(base10)1.6
3x log(base10)2 = log(base10)1.6 (divide both sides by log(base10)2 to get 3x on one side by itself)
3x = 0.2041199827/0.3010299957
3x = 0.6578150531 (divide both sides by 3 to get x by itself)
x = 0.22 (rounded to 2 decimal places)

hope this helps ur 3rd question...i will try to work out the other ones for you soon as I am a little busy with my maths assignment due soon as well...

here's the other one...
log(base1/3)(x^2 + x) – log(base1/3)(x^2 – x) = -1
log(base1/3)[(x^2 + x)/(x^2-x)] = -1 (combine the logs)
(1/3)^-1 = [(x^2 + x)/(x^2-x)] (use the log rule: log(base a) b = x ---> a^x = b
3 = [(x^2 + x)/(x^2-x)] [(1/3)^-1 = 3]

So just work this out from here, as I don’t have the time to work it out fully, unless u give me ur email address coz it takes YEARS to type it out...i really have 2 write out all the steps and and scan them in...which would be much quicker...anyway, hope this helps for the 6th question! =D

Answer 4

ZeroCarbonImpact,

I want to know that too. Actually, which question in mathematics is not doing someone else's homework?

Answer 5

You want us to do all your homework?