Help Please!!!...logs!?!?!?!?

Question

ok here are some problems that I'm completely stuck on..can someone help ...if you know atleast one that would be great I just need help on these.. Thanks

x= the variable not multiplication for those who asked on my other question.

Here I have to use properties of logs to condense the logs to a single term:

1/2 log base 10 x + 3 log base 10 (x+1)


Here I have to expand the following expressions using logs:

√3x-5
-------- (All of that goes under the square root sign)
7


4x √10-3x

(Only the 10-3x are under the square root sign)


Here I have to solve the exponential equation using log properties:

2 log base 5 3x = 4



log base 3 x + log base 3 {(x^2) - 8 } = log base 3 8x

Answer 1

# 1.
1/2 log (base 10) x = log (base 10) x^(½)

3 log (base 10) (x+1) = log (base 10) (x+1)³

Since the logs are added, it is implied that we are taking the log of a product, in this case [x^(½)](x+1)³. Note, x > 0, otherwise we are taking the square root of a negative number.

So 1/2 log (base 10) x + 3 log (base 10) (x+1) = log (base 10) {[x^(½)](x+1)³}.


#2.
√3x-5
-------- (All of that goes under the square root sign)
7

I assume here you mean the entire expression goes under a radical sign and that 3x - 5 also go under their own radical sign. If they don't, all bets are off.

Allow me to rewrite this as [(3x-5)^½ / 7]^½.

log [(3x-5)^½ / 7]^½ = ½ [log (3x-5)^½ - log 7] =
½ [log (3x-5)^½] - ½ log 7 = ½[½ log (3x - 5)] - ½ log 7 =
¼ log (3x - 5) - ½ log 7.

Note that 3x - 5 > 0. So x > 5/3, otherwise we are taking the square root of a negative number. Also, note that (3x - 5) is not further reducible, because it is neither the product nor quotient of a number.


#3.
log (4x √10-3x) = log 4 + log x + ½ log (10 - 3x).

This is a product of three numbers, so we add their separate logs. (10 - 3x) is not further reducible, since it is neither a product nor a quotient. Also, (10 - 3x). must be greater than 0. Therefore x < 10 / 3 because we cannot take the log of a negative number.


#4.
2 log (base 5) 3x = 4

First divide both sides by 2 to get:

log (base 5) 3x = 2.

That implies 3x = 5² = 25 ---> x = 25 / 3.

Check by substituting 25 / 3 back into the original equation.


#5.
log (base 3) x + log (base 3) {(x^2) - 8 } = log (base 3) 8x
log (base 3) x + log (base 3) {(x^2) - 8 } = log (base 3) 8 + log (base 3) x

We can subtract log (base 3) x from both sides, leaving us with:

log (base 3) {(x^2) - 8 } = log (base 3) 8.

The above implies (x^2) - 8 = 8. So (x^2) = 8 + 8 = 16.

Here we have to be careful. Ordinarily we could use x = ±sq rt (16) for our solution. But because we have used x alone in other parts of our equation, we are compelled to use the positive value only, because we cannot take the log of a negative number. Therefore, x = 4 is our solution in this case.

Answer 2

1/2 log base 10 x + 3 log base 10 (x+1)
= log(x+3)^1/2 + log (x+1)^3
=log[(x+3)^1/2*(X+1)^3]

sqrt[(3x-5)/7]^1/2
=[sqrt (3x-5)]/sqrt(7)
= 1/2log(3x-5) -1/2log 7

4x √10-3x
= log 4x + 1/2log(10-3x)

2 log base 5 3x = 4
log_5 (3x)^2 = 4
5^log_5 9x^2 = 5^4
9x^2 = 5^4
3x = 5^2
x = 25/3

log base 3 x + log base 3 {(x^2) - 8 } = log base 3 8x
= log_3 x*(x^2-8) = log_3 8x
x^3 -8x = 8x
x^3 = 16x
x = 2cuberoot(2)