SOLUTION: PLEASE solve these problems PLEASE !! log(3)5+log(3)x=log(3)10 log(1/7)x=-1 2ln3x+1=5 log(3)0.1+2log(3)x=log(3)2+log(3)5 log(b)9=2 1/4log(2)16+1/2log(2)49=log(2)2x 3^2m=16

Question 475188: PLEASE solve these problems PLEASE !!
log(3)5+log(3)x=log(3)10
log(1/7)x=-1
2ln3x+1=5
log(3)0.1+2log(3)x=log(3)2+log(3)5
log(b)9=2
1/4log(2)16+1/2log(2)49=log(2)2x
3^2m=16^-1
4^ex-5=1
THESE ARE THE ANSWER CHOICES :
2 , 5 , 14 , -2 , 3.2147 , 10 , 7 , 1.0986 , 0.5537 , 4 , 2.4630 , 3
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
log(3)5+log(3)x=log(3)10

Adding logs --> multiply

5x = 10
x = 2
-----------------
log(1/7)x=-1
Is that log of (x/7) ? or log or x*log(1/7) ?
-------------------
2ln3x+1=5
2ln(3x) = 4
ln(3x) = 2
3x = e^2
x = e^2/3
----------
log(3)0.1+2log(3)x=log(3)2+log(3)5
Similar to the 1st one.
--------------
log(b)9=2
b^2 = 9
b = 3
---------------
1/4log(2)16+1/2log(2)49=log(2)2x
Similar to the 1st one.
----------------------
3^2m=16^-1
Is 2 the exponent? or 2m?
----------------
4^ex-5=1
4^ex = 6x
ex = 6^(1/4)
x = 6^(1/4)/e

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