The number of solutions of log 4x-1-log 2x-3 isA- 3B- 1C- 2D- 0

The correct option is B 1log_4 (x 1)=log_2(x 3) x 1 ≥ 0 and x 3≥ 0 ∴ x ≥ 3 1/2 log_2(x 1)=log_2(x 3) (x 1)=(x 3)^2 x 1=x^2+9 6x x^2 7x+10=0 (x 2)(x 5)=0 x=2 o ...

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Byju's Answer

Standard IX

Mathematics

Logarithmic Inequality

The number of...

Question

The number of solutions of $l o g_{4} (x - 1) = l o g_{2} (x - 3)$ is

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Solution

The correct option is B 1
$l o g_{4} (x - 1) = l o g_{2} (x - 3)$
$x - 1 \geq 0 a n d x - 3 \geq 0 ∴ x \geq 3$
$\frac{1}{2} l o g_{2} (x - 1) = l o g_{2} (x - 3)$
$(x - 1) = (x - 3)^{2}$
$x - 1 = x^{2} + 9 - 6 x$
$x^{2} - 7 x + 10 = 0$
$(x - 2) (x - 5) = 0$
$x = 2 o r x = 5$
$x = 2$ is not accept able $(a s x s h o u l d b e \geq 3)$
so, no.of solution is = 1

Suggest Corrections

The number of solutions of log4(x−1)=log2(x−3) is

The correct option is B 1log4 (x−1)=log2(x−3) x−1≥0 and x−3≥0∴x≥3 12log2(x−1)=log2(x−3) (x−1)=(x−3)2 x−1=x2+9−6x x2−7x+10=0 (x−2)(x−5)=0 x=2 or x=5 x=2 is not accept able (asxshouldbe≥3) so, no.of solution is = 1

The number of solutions of $l o g_{4} (x - 1) = l o g_{2} (x - 3)$ is