How to Solve Logarithmic and Exponential Equations with Different Bases: Step-by-Step Tutorial

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PreMath

PreMath

Күн бұрын

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-uv9fq
@BobBob-uv9fq 3 жыл бұрын
I like how he says “huuuuullllo again everyone “
@Matt-iy7pr
@Matt-iy7pr 2 жыл бұрын
Thank u for vid! It help me practice logarithm
@basilrex4105
@basilrex4105 6 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
@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-uv9fq
@BobBob-uv9fq 3 жыл бұрын
It actually makes me chuckle watching this ,how cute
@keithedwards2841
@keithedwards2841 6 жыл бұрын
Excellent teaching
@PreMath
@PreMath 6 жыл бұрын
Thanks Keith Edwards for the feedback. Your input is always appreciated. Kind regards
@kibeterick5776
@kibeterick5776 2 жыл бұрын
This is amazing
@keithedwards2841
@keithedwards2841 6 жыл бұрын
This teaching makes math easy
@PreMath
@PreMath 6 жыл бұрын
Thanks again Keith Edwards. Kind regards
@ЈеленаЦавнић
@ЈеленаЦавнић 4 жыл бұрын
He made it harder...He always ue this.metod and this metod is long and hardest....
@AmirgabYT2185
@AmirgabYT2185 8 ай бұрын
x=2
@jemanishajones9707
@jemanishajones9707 3 жыл бұрын
I watched and liked it
@bongthomreviewkh563
@bongthomreviewkh563 3 жыл бұрын
thanks
@ganeshdas3174
@ganeshdas3174 2 жыл бұрын
The given expression comes to x=2 after change' of base law, cancellation and exponential form
@devondevon4366
@devondevon4366 4 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
@PreMath
@PreMath 4 жыл бұрын
Thanks for the feedback. Kind regards :)
@itsahmd295
@itsahmd295 4 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
@adgf1x
@adgf1x Жыл бұрын
=>5/3+log(x-1)base2=5/3=>x-1=2^0=1=>x=2ans
@calipsoceo7812
@calipsoceo7812 4 жыл бұрын
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