How to Solve Exponential Equations with Logarithmic Exponents Using Neat and Simple Tricks

[VolksdeutscheSS]

Excellent exponential equation and solution, Sir. One suggestion: In the beginning it is unclear whether the final term is log base 5 or whether it is log base 10 with argument 5.

[ananeakwasi]

I need help

Log(3exponent x-1 +3 exponent 2-x)=log7+log4/log2

[brucemaclennan]

When you first see this equation it appears a little muddling in
Its youtube window.
In the third term the 5 seems to appears to be on the same line
As the base numbers in the two prior expressions suggesting
There is an omission ...the number to be referred to the base 5
Has been omitted and there is possibly a typo here

[owlsmath]

awesome and very thorough. I got it wrong because i lost a 1 somewhere and it threw the whole thing off. My bad

[pinklady]

I love your channel with challenging maths.

[devondevon]

Answer x=2 and x=3
49^log7 (x-1) - 5^log 5 3 x + 10^log 5 =0
7^2 log 7 (x-1) - 5^log 5 3x + 10^log 5=0
7 log 7 (x-1)^2 - 5 log 5 3x + 10^log 5=0
(x-1)^2 - 3x + 5=0
x^2-2x+1-3x +5=0
x^2-5x+6=0
(x-3)(x-2)=0
x=3 or x= 2 answer
let check using x=2 then 49^0 - 5^ log 5 6 + 10^log5=0
1-6 +5=0

[MathsKaDar]

Sir thumbnail dekhlo LHS ki last term k exponent k logarithmic ka bade lg rha h 5 instead of argument

[waltjhall]

5:49 You can also use 6 and 1 (x-6)(x+1) = 0

[adgfx]

(x-1)^2-3x+5=x^2-5x+6=0=>(x-3)(x-2)=0=>x=2, 3.ans

[chariteishimwe]

Pre math thanks a lot for your explanations

[theodoresweger]

I simply plug it into my hp prime, problem solve