1. ## Textbook questions i had trouble with

Hey, these are questions i had trouble with

Solve the following for x,

b. log e (5x) - log e (3-2x) +1
c. Log e (x) + log e (3x+1)=1
d. (8e^-x) - e^x =2

Simplify,
a. 1+ log10 a - (1/3)log10 b
b. log2 x -2log2 y + log2(xy^2)

Evaluate,
a. log2 square root 2 + log2 1 +2log2 2

Thanks

2. Originally Posted by chaneliman
Hey, these are questions i had trouble with

Solve the following for x,
a. 2 log e (x+5) =6
b. log e (5x) - log e (3-2x) +1
c. Log e (x) + log e (3x+1)=1
d. (8e^-x) - e^x =2

Simplify,
a. 1+ log10 a - (1/3)log10 b
b. log2 x -2log2 y + log2(xy^2)

Evaluate,
a. log2 square root 2 + log2 1 +2log2 2
b. 2log10 5 + 2log10 2 +1

Thanks
Here are a few c) from the first group
$\displaystyle \log_e(x)+\log_e(3x+1)=1$ rewrite using prop of logs

$\displaystyle \log_e(x[3x+1])=1 \iff x(3x+1)=e \iff 3x^2+x-e=0$

using the quadratic formula we get

$\displaystyle x=\frac{-1 \pm\sqrt{1-4(4)(-e)}}{2(3)}=\frac{-1 \pm \sqrt{1+16e}}{6}$

b) from the second group

$\displaystyle \log_2(x)-2\log_2(y)+\log_2(xy^2) \iff \log_2(x)-\log_2(y^2)+\log(xy^2)$

$\displaystyle \log_2(\frac{x}{y^2})+\log_2(xy^2)=\log_2(\frac{x} {y^2}\cdot xy^2)=\log_2(x^2)$

3. yea i get those 2 now , but i'm still a little confused with the others