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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 -

By definition, blogbN = 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 -

Thanks a lot!

• Logs - correction -

#7. Actually, x = -2 needs to be discarded, since log(x-2) is not defined there.

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