Question

log4 to the base x-1=1+logx-1 to the base 2




please ans fast i will mark you as brainliest ​

Answer 1

Answer:

log 1 base 2= 0

as2^0=1

now, log (x-1) +log(x+1) to the base 2

log (x-1) +log(x+1)=0

log ( x^2-1)=0

10^0=x^2-1

1+1=x^2

x=√2 is your ans sister

MARK ME AS A BRAINLIST

Answer 2

Solution:

$log_{x - 1}(4) = 1 + log_{2}(x - 1)$

Using the base change property and equating the base equal to 2 for both log

$\frac{ log_{2}(4) }{ log_{2}(x - 1) } = 1 + log_{2}(x - 1)$

CROSS MULTIPLYING

$2 = log_{2}(x - 1) + log_{2}(x - 1) ^{2}$

LET

$log_{2}(x - 1) = y$

${y}^{2} + y - 2 = 0$

$(y - 1)(y + 2) = 0$

AS VALUE OF log cannot be negative

$so \\ y = 1 \\ i.e \\ log_{2}(x - 1) = 1 \\ x - 1 = {2}^{1} = 2 \\ x = 3$

SO THE ANSWER IS x=3

thanks and Mark me as the brainliest.