Posted by **Jessie** on Thursday, December 19, 2013 at 11:42pm.

1. Evaluate 4^(log base4 64) + 10^(log100)

2. Write 1+log(base2)x^3 as a single logarithm

3. Write log(base b)√(x^3 y z^6)

4. Solve log(base 2)x-log(base 2)6=log(base 2)5+2log(base 2)3

5. Solve 3^(2x) = 9(81^x)

6. Solve 3^(2x)=7^(3x-1). Round answer to two decimal places.

7. Solve log(x+3) + log(x-2) = log(3x+2)

- Advanced Functions/ Precalculus Log -
**Steve**, Friday, December 20, 2013 at 12:24am
By definition, b^{logbN} = N

#1. 64+100 = 164

#2. 1 = log2, so you have

log2+log(x^3) = log(2x^3)

#3. 1/2 (3logx + logy + 6logz)

#4. supressing the "base 2" for readability, we have

logx - log6 = log5+2log3

log(x/6) = log(5*3^2)

x/6 = 45

x = 270

Actually, since all the logs are the same base, the base doesn't really matter.

#5. Since 9=3^2 and 81=3^4, you have

3^(2x) = 3^2 * 3^4x

3^2x = 3^(2+4x)

2x = 2+4x

x = -1

Check: 3^-2 = 1/9

9(81^-1) = 9/81 = 1/9

#6.

3^(2x) = 7^(3x-1)

2x log3 = (3x-1) log7

2x/(3x-1) = log7/log3

x = 0.53

#7.

log (x+3)(x-2) = log (3x+2)

(x+3)(x-2) = (3x+2)

x = -2 or 4

- Advanced Functions/ Precalculus Log -
**Jessie**, Friday, December 20, 2013 at 2:55am
Thanks a lot!

- Logs - correction -
**Steve**, Friday, December 20, 2013 at 12:14pm
#7. Actually, x = -2 needs to be discarded, since log(x-2) is not defined there.

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