Question

use natural logarithms to solve the equation. round to the nearest thousandth 3e^(2x) +2=28 WILL GIVE MEDALS

Answer 1

I put it in my TI-89 calculator and got 165.794

Does that sound right to you?

Answer 2

no my choices are
A. 1.0797
B. 0.4689
C. 0.9261
D. 2.1595

Answer 3

\[3e^{2x}+2=28\]Subtract 2 then divide by 3.\[e^{2x}=\frac{26}{3}\]Take the natural logarithm of both sides.\[\ln(e^{2x})=\ln(\frac{26}{3})\]ln and e cancel out.\[2x=\ln(\frac{26}{3})\]Divide by 2.\[x=\frac{1}{2}\ln(\frac{26}{3})\]Put this in a calculator.\[x=1.0797\]So it's A.

Answer 4

THANK YOU

Answer 5

can you help me with another one

Answer 6

sure

Answer 7

write the expression as a single logarithm 4 log^by 3log^bx

Answer 8

Sorry, I had to leave for something. I'll still do it in case you're still here. I'm assuming the ^ mean they're just arguments to the log.\[4\log(by)+3\log(bx)\]\[\log[(by)^4]+\log[(bx)^3]\]\[\log[(by)^4(bx)^3]\]\[\log(b^7y^4x^3)\]

Answer 9

I also assumed that there was supposed to be a plus sign between them because that makes much more sense.

Answer 10

@bangajr can you help i had to go to tennis practice sorry my choices are
\[A. \log _{b}(y ^{4}x ^{3})\]
\[B.\log _{b}(y^{4}+x ^{3})\]
\[C.\log _{b}(yx ^{4+3})\]
\[(4+3)\log _{b}(y+x)\]