How to solve a complex logarithm question

[@PowerLearnScienceandMathsTV]

Please leave a comment behind on how you see the solution.

[@wisdomfonore]

Great 👌

[@karpholmes6942]

Log2(4) = m
Thus
Log2(x-1)= log4(x+5)
Can be shown as
Log4(x-1)^m= log4(x+5)
m =2 and remove common logs
(X-1)^2=x+5
X^2 -2x +1 = x+5
X^2 -3x -4=0
Factor or quadratic formula. I’ll will do the first
(X-4)(x+1)= 0
X= 4 and 1 10:39
Log2(-2) is undefined and 4 isn’t.
3=+sqrt9
4=x

[@Paul-222]

Log base 2 (x-1) = log base 4 (x-1)^2. Remove the logs, solve the quadratic.

[@PowerLearnScienceandMathsTV]

Great. That works too. Thanks.

[@MartinPerez-oz1nk]

THANKS PROFESOR!!!!, VERY INTERESTING!!!!!!

[@PowerLearnScienceandMathsTV]

Thanks for your comment sir.
Pls don't forget to subscribe😊

[@jan-willemreens9010]

... Good day to you, Of course there are multiple ways to solve such equations, and I'm more than 100% sure (lol) you know that too! Given: LOG(b2)(X - 1) = LOG(b4)(X + 5) [ Applying the Change of Base formula and setting up the restriction(s) for possible X's ] ... X - 1 > 0 and X + 5 > 0 ... X > 1 and X > - 5 so the combined restriction is X > 1. LOG(b2)(X - 1) = LN(X - 1)/LN(2) and LOG(b4)(X + 5) = LN(X + 5)/LN(4) = LN(X + 5)/(2*LN(2)) = (1/2)LN(X + 5)/LN(2) = LN((X + 5)^(1/2))/LN(2) [ Equating both changed log expressions (LN) ] ... LN(X - 1)/LN(2) = LN((X + 5)^(1/2))/LN(2) ... LN(X - 1) = LN((X + 5)^(1/2)) ... X - 1 = (X + 5)^(1/2) ... (X - 1)^2 = X + 5 [Skipping some obvious simplifying steps ] ... X^2 - 3X - 4 = 0 ... (X - 4)(X + 1) = 0 ... X1 = 4 v X2 = - 1 [ Respecting the inequality restriction X > 1 ] ... X1 = 4 is the only valid solution for the original equation ( I also checked this solution in the original equation, just to be sure ) ... finally obtaining S = { 4 } ... it is always interesting/educational to watch different people solving the same problem, and I thank you for your more elaborate presentation and in general for your math efforts ... Best regards, Jan-W

[@PowerLearnScienceandMathsTV]

Great comment. You are great teacher. Thanks

[@honestadministrator]

log (x+5) to the base 4
= log (x+5) to the base 2
/ log 4 to the base 2
= log (x+5) to the base 2 / 2
Hereby given expression results in
(x-1) ^2 = ( x +5)
or x^2 - 3 x - 4 = 0 i.e. (x-4) (x+1) = 0
or x = 4, -1
For the given equation only feasible solution is x= 4

[@PowerLearnScienceandMathsTV]

That's great. You used change of base . Thanks for your comment. Pls don't forget to subscribe

[@bobwineland9936]

From logbase10 of (X-1)^2 =logbase 10(X+5), and since the bases are the same then the exponents are equal to each other. Seems like a faster way to solve.

[@PowerLearnScienceandMathsTV]

Good point. You are right sir. Thank you

[@hellohabibi1]

x=4?

[@shadmanhasan4205]

Answer: f(4) = ~1.585

[@ErmiyasTsegu]

I love you!!!

[@trojanleo123]

x = {-1,4} but -1 is extraneous, so x = 4 is the only real root.

[@muhammademran2464]

Don't need to mention (write) the base if it's 10 throughout
Log(x-1)/log2 = log (x+5)/log4
2Log(x-1) = log (x+5)
X^2 -2x + 1 = x +5
X^2 - 3x -4 =0
X=4, x=-1

[@PowerLearnScienceandMathsTV]

Thanks

[@studieswithak3383]

Bro there's a simple method to solve this by looking
log 4(x+5)=1/2log2(x+5)

[@PowerLearnScienceandMathsTV]

You are right bro. Wanted to explore the other way of solving that. Thanks for your valuable contribution.

[@Heaven_Vs_Hel]

x-1=√(x+5), because log(x+5) to the base 4 is log[√(x+5)] to the base 2, where x>1. 😀😉

[@NurHadi-qf9kl]

2x-2=x-5
.x=-3

[@NurHadi-qf9kl]

O...x=7

[@anandamondal6092]

X=7