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

[kennethokoro]

Thanks for taking the time to really teach this God bless you the more with wisdom to teach others in JESUS name.Amen.

[Matt-iypr]

Thank u for vid! It help me practice logarithm

[basilrex]

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

[BobBob-uvfq]

It actually makes me chuckle watching this, how cute

[ganeshdas]

The given expression comes to x=2 after change' of base law, cancellation and exponential form

[devondevon]

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

[itsahmd]

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