Hivaclibtibcharkwa 𝗕𝗶𝗼𝗹𝗼𝗴𝘆 𝗧𝘂𝘁𝗼𝗿 Joined Jan 17, 2021 Messages 1,539 Gender Male HSC 2021 Sep 7, 2022 #1
C cossine Active Member Joined Jul 24, 2020 Messages 370 Gender Male HSC 2017 Sep 7, 2022 #2 Hivaclibtibcharkwa said: View attachment 36204View attachment 36205 Click to expand... If you exponentiate both sides. You would get: e^(ln(x+1)+ln(2x+1)) = e^ln(4).
Hivaclibtibcharkwa said: View attachment 36204View attachment 36205 Click to expand... If you exponentiate both sides. You would get: e^(ln(x+1)+ln(2x+1)) = e^ln(4).
J jks22 Member Joined Jan 18, 2022 Messages 74 Gender Male HSC 2022 Sep 7, 2022 #3 cossine said: If you exponentiate both sides. You would get: e^(ln(x+1)+ln(2x+1)) = e^ln(4). Click to expand... Yeah and if you split it, it'll be e^ln(x+1) x e^ln(2x+1), not addition
cossine said: If you exponentiate both sides. You would get: e^(ln(x+1)+ln(2x+1)) = e^ln(4). Click to expand... Yeah and if you split it, it'll be e^ln(x+1) x e^ln(2x+1), not addition
Hivaclibtibcharkwa 𝗕𝗶𝗼𝗹𝗼𝗴𝘆 𝗧𝘂𝘁𝗼𝗿 Joined Jan 17, 2021 Messages 1,539 Gender Male HSC 2021 Sep 7, 2022 #4 cossine said: If you exponentiate both sides. You would get: e^(ln(x+1)+ln(2x+1)) = e^ln(4). Click to expand... Thank you! So the whole side of the equation becomes to the power of e
cossine said: If you exponentiate both sides. You would get: e^(ln(x+1)+ln(2x+1)) = e^ln(4). Click to expand... Thank you! So the whole side of the equation becomes to the power of e
0 011235 Active Member Joined Mar 6, 2021 Messages 136 Gender Male HSC 2023 Sep 7, 2022 #5 Correct, you have to change both sides of the equation the same way, not term by term
D Drongoski Well-Known Member Joined Feb 22, 2009 Messages 4,090 Gender Male HSC N/A Sep 7, 2022 #6 You don't need to exponentiate. ln(x+1) + ln(2x+1) = ln 4 .: ln(x+1)(2x+1) = ln 4 .: (x+1)(2x+1) = 4 Then solve. Remember the argument of ln must be positive; we need x+1 > 0 and 2x+1 > 0.
You don't need to exponentiate. ln(x+1) + ln(2x+1) = ln 4 .: ln(x+1)(2x+1) = ln 4 .: (x+1)(2x+1) = 4 Then solve. Remember the argument of ln must be positive; we need x+1 > 0 and 2x+1 > 0.