how to solve log base 2 (x-1) = log base 4 (x+1)?

We can rewrite the expression as..

log(x - 1)/log(2) = log(x + 1)/log(4)

log(x - 1)/log(2) = log(x + 1)/(log(2²))

log(x - 1)/log(2) = log(x + 1)/(2log(2))

log(x - 1)/log(2) = log(x + 1)/log(2)

log_2 (x - 1) = log_2 √(x + 1)

Then..

(x - 1)² = √(x + 1)²

x² - 2x + 1 = x + 1

x² - 3x = 0

x(x - 3) = 0

Finally...

x = 0 and x - 3 = 0

x = {0, 3}

Since x = 0 makes the log expression imaginary, reject that. Hence, x = 3.

I hope this helps!

Convert all logs to the same base using

Log (base a) (b) = (Log b)(Log a)

The equation becomes

Log(x-1)/Log2 = Log(x+1)/Log4

Log(x-1)/Log2 = Log(x+1)/(2*Log2)

2* Log2 * Log(x-1) = Log(x+1) * Log2

2*Log(x-1) = Log(x+1)

Log(x-1)^2 = Log(x+1)

(x-1)^2 = x+1

x^2-2x+1 = x+1

x^2-3x = 0

x(x-3) = 0

x = 0,3

Now we can't take x=0 as a solution, as the in the initial problem we would then have Log (0-1) = Log(-1) which is not defined. So the only allowed solution is

x=3

log base 2 (x-1) = log base 4 (x+1) = log base 4 (2) log base 2 (x + 1)

whence (x -- 1)^2 = (x + 1) giving x = 3, 0

first get this formula...... ln base m (x) = ln base n (x) /ln base n (m)

to simplify your question its good to change everything in base e or 10... i am doing with 10, you try it by taking e..

so with base 10 you will have ... log base10 (x-1)/log base 10 (2) = log base 10 (x+1)/log base10 (4)

As now base are same.. we can write as ==...... log(x-1)/log (x+1) = [log (2)/log(4)]

= log(x-1)/log (x+1) = 0.5

=log(x-1) = 0.5*log (x+1)

= log(x-1) = log (x+1)^1/2

taking Antilog on both side..(as log and antilog cancels each other)

= (x-1) = (x+1)^1/2

taking square on both sides

= (x-1)^2 = (x+1)

= x^2 -2x +1 = x+1

= x^2 -2x +1 - x-1 =0 which implies x^2-3x = 0 means x(x-3)=0

So the answer is x= 0 or (x-3)=0 that is x=3.......

log base 2 (x-1) = log base 4 (x+1)

log base 2 (x-1) = log base 2 raised to power 2 (x+1)

log base 2 (x-1) = (1/2)log base 2 (x+1)

log base 2 (x-1) = log base 2 [(x+1)raised to power 1/2]

since two numbers give the same logarithm so they must be equal

x-1=(x+1)raised to power 1/2

(x-1)raised to power 2=x+1 ;on squaring both sides

x^2+1-2x=x+1

x^2-3x=0

x(x-3)=0

so x=0 or x=3

but x cannot be zero as LHS's (x-1) will become less than 0

so x=3 is the required answer

Explanation:-

rules used are

log base p raised to power q (a raised to power b)=(b/q)log base p (a)

and

n log base a (p)=log base a (p raised to power n)

log base 2 (x-1) = 1/2log base 2 (x+1)

Using the identity- log a^b (C)= 1/blog base a (C)

thus you get,

log base 2 (x-1)= log base 2 (underoot of x+1)

removing the log

x-1= (x+1)^1/2

Squaring both sides

(x-1)^2= x+1

x^2 + 1 - 2x = x +1

x^2 - 3x= 0

x(x-3) = 0

hence, x= 0 or x= 3

Good luck!

.Q. Solve for x : log base2 (x-1) = log base4 (x+1)?

Ans : Given : log base2 (x-1) = log base4 (x+1)

Apply : log baseX (Y) = log(Y) / log(X)

=> log base2 (x-1) =log base4 (x+1)

=> [log(x-1)] / log(2) =[log (x+1)] / log(4)

=> [log(x-1)] / log(2) =[log (x+1)] / log{(2^2)}

Apply : log(x^n) = n*log(x)

Therefore : [log(x-1)] / log(2) =[log (x+1)] / log{(2^2)}

=> [log(x-1)] / log(2) =[log (x+1)] / [2*log{(2)}]

Cancel log(2) both sides :

Therefore : [log(x-1)] =[log (x+1)] / [2]

=> 2*[log(x-1)] =[log (x+1)]

Apply : n*log(x) = log(x^n)

Therefore : log {(x-1)^2}=[log (x+1)]

=> log {(x-1)^2} -[log (x+1)] = 0

Apply : log(a) - log(b) = log(a/b)

Therefore : log {(x-1)^2} -[log (x+1)] = 0

=>log {(x-1)^2 / (x+1) } = 0

Take Exponential on both sides :

=>e^ [log {(x-1)^2 / (x+1) } ]= e^(0)

=> {(x-1)^2 / (x+1) } = 1

=> (x-1)^2 = (x+1)

=> {x^2+1-2x} = (x+1)

=> {x^2-3x}=0

=> x(x-3) = 0

=> x = 0 or x = 3

The value of x : 0 or 3

I wish you will appreciate my answer.