How to Solve Logarithmic and Exponential Equations with Different Bases and Check Answers for Extraneous Solutions. Also Learn other Essential Logarithmic Rules. Step-by-Step Tutorial by PreMath.com
Пікірлер: 21
@BobBob-uv9fq3 жыл бұрын
I like how he says “huuuuullllo again everyone “
@Matt-iy7pr2 жыл бұрын
Thank u for vid! It help me practice logarithm
@basilrex41056 жыл бұрын
In this particular equation since + 5/3 appears on both the right and left side, the value of the log is 0 which means x - 1 =1 and x = 2
@kennethokoro6817 Жыл бұрын
Thanks for taking the time to really teach this God bless you the more with wisdom to teach others in JESUS name.Amen.
@BobBob-uv9fq3 жыл бұрын
It actually makes me chuckle watching this ,how cute
@keithedwards28416 жыл бұрын
Excellent teaching
@PreMath6 жыл бұрын
Thanks Keith Edwards for the feedback. Your input is always appreciated. Kind regards
@kibeterick57762 жыл бұрын
This is amazing
@keithedwards28416 жыл бұрын
This teaching makes math easy
@PreMath6 жыл бұрын
Thanks again Keith Edwards. Kind regards
@ЈеленаЦавнић4 жыл бұрын
He made it harder...He always ue this.metod and this metod is long and hardest....
@AmirgabYT21858 ай бұрын
x=2
@jemanishajones97073 жыл бұрын
I watched and liked it
@bongthomreviewkh5633 жыл бұрын
thanks
@ganeshdas31742 жыл бұрын
The given expression comes to x=2 after change' of base law, cancellation and exponential form
@devondevon43664 жыл бұрын
log 8 32 + log 2 (x-1) = 5/3 log 8 32 = log 2 32 ______________ = 5/3 log 2 8 So log 2 (x-1) = 5/3 - 5/3 = 0 So log 2 (x-1) = 0 Therefore x= 2 since 2^0= 1 and for x-1 to equal 1 then x , must = 2 as x-1= 1 , x must= 1+1= 2 Answer x =2 note that since log 8 32 also =5/3, then log 2 (x-1)=0, therefore the argument '(x-1)' must = 1 therefore x must =2, so answer again 2
@PreMath4 жыл бұрын
Thanks for the feedback. Kind regards :)
@itsahmd2954 жыл бұрын
No need. 8 min for this. 8^y=32 2^3y=2^5 3y=5 y=5/3 then substitute. Log2(x-1)=5/3 - 5/3 log2(x-1)=0 2^0= x-1 x=1+1 x=2