Logarithmic Equations | how to evaluate logarithms without a calculator.

Logarithm is one of the most interesting topics in mathematics.
In this video tutorial, I, will guide you on how to evaluate logarithms without the use of a calculator by simplifying the logarithmic expression through the use of some of the rules or properties of logarithms.

Some of the rare properties of logarithms applied here are: the reciprocal law of logarithms, the power law of logarithms, change of base law of logarithms, among others.

Watch from the beginning to the end without skipping any parts of this video for the full details and also share with those that need this knowledge using the video link below.

Much love 💖💖😍😍

#logarithm #log #simplify

[MarcosSantos-scdf]

congratulations on the didactics in teaching teacher

[tygrataps]

Very interesting. I ended up taking the route with change of base instead. I hadn't come across the log base/exponent idea rule before. Learning something new all the time. Thank you!

[AlaaAlaa-erld]

4:32
so u must not equal zero
then x must not equal 1
if x = 1 then a is undefined

[evic]

Easy way: log2(x) = 2log4(x): 1/2*log4(x) = loga(x) => log16(x) = loga(x)

[nagarajahshiremagalore]

In all your problem sessions, i find there is a lot of scope to reduce the time of running the video. Pl impliment.

[AlaaAlaa-erld]

because log_x y = 1 / log_y x
》1 / (log_x 4) ^ 2 = 1 / log_x 2 • 1 / log_x a
》(log_x 4) ^ 2 = log_x 2 • log_x a
》(2 • log_x 2) ^ 2 = log_x 2 • log_x a
》4 • log_x 2 = log_x a
》 a = 2 ^ 4 = 16

😅😂

[budisuhuyanli]

n^6-n^3= 2
ada 6 solusi untuk n,

Jika
n^3 = a
Maka, a^2-a= 2

a^2 - a - 2 = 0
(a-2)(a+1)=0
a=2 atau a=-1
Jadi
n^3= 2 atau n^3= -1

Jika sumbu Y adalah imaginer
Dan sumbu X adalah Real

Maka:
n^3= 2
n1= akar3(2)*[Cos 0° - i*Sin 0°]
n2= akar3(2)*[Cos 120° - i*Sin 120°]
n3= akar3(2)*[Cos 240° - i*Sin 240°]

n^3= -1
n4= [Cos 60° - i*Sin 60°]
n5= [Cos 180° - i*Sin 180°]
n6= [Cos 300° - i*Sin 300°]

Disederhanakan maka:
n1= akar3(2)

n2= -akar3(2) * [i*(√3)+1]/2

n3= akar3(2) * [i*(√3)-1]/2

n4= [1-i*(√3)]/2

n5= -1

n6= [1+i*(√3)]/2

[lzuluaga]

Por qué pierdes el tiempo reemplazando por U? Cancela de una.

[sans]

(log(4) -> x)^2 = log(2)->x * log(a)->x
=>

/(lnx*lnx) on both sides
1/(ln4*ln4)=1/(ln2*lna)
ln4*ln4=ln2*lna
raise e to both sides
(e^ln4)^ln4=(e^lna)^ln2
4^ln4=a^ln2
raise both sides to 1/ln2
a=4^(ln4/ln2)
edit: wait what-
edit 2: oh wait
ln4=ln(2^2)=2ln2
2ln2/ln2=2
a=4^2=16