Question

log2 (x - 1) = 1 + log2 (2x - 3)

plz tell tomorrow test . and 2 is base ​

Answer 1

Question :-

Solve for x :

$\rm \: log_{2}(x - 1) = 1 + log_{2}(2x - 3)$

$\large\underline{\sf{Solution-}}$

Given Logarithmic equation is

$\rm \: log_{2}(x - 1) = 1 + log_{2}(2x - 3)$

Before start the solution, let first define the domain.

So, for given equation to be defined,

$\rm \: x - 1 > 0 \: \: and \: \: 2x - 3 > 0$

$\rm \: x > 1 \: \: and \: \: 2x > 3$

$\rm \: x > 1 \: \: and \: \: x > \dfrac{3}{2}$

$\rm\implies \: \: x > \dfrac{3}{2}$

Now, Consider

$\rm \: log_{2}(x - 1) = 1 + log_{2}(2x - 3)$

We know,

$\boxed{\sf{  \:\rm \: log_{x}(x) = 1 \: \: }}$

So, using this result, the above expression can be rewritten as

$\rm \: log_{2}(x - 1) = log_{2}(2) + log_{2}(2x - 3)$

We know,

$\boxed{\sf{  \:\rm \: log_{a}(x) + log_{a}(y) = log_{a}(xy) \: \: }}$

So, using this result, we get

$\rm \: log_{2}(x - 1) = log_{2}(2(2x - 3))$

$\rm \: log_{2}(x - 1) = log_{2}(4x - 6)$

$\rm \: x - 1 = 4x - 6$

$\rm \: x - 4x = 1 - 6$

$\rm \: - 3x = - 5$

$\rm\implies \:x = \dfrac{5}{3} \: \in \: \bigg(\dfrac{3}{2} , \: \infty \bigg)$

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Additional Information :-

$\boxed{\sf{  \:\rm \: log_{a}(x) - log_{a}(y) = log_{a} \bigg(\frac{x}{y}\bigg) \: \: }}$

$\boxed{\sf{  \:\rm \: log_{a}( {x}^{y} ) = y \: log_{a}(x) \: \: }}$

$\boxed{\sf{  \:\rm \: log_{x}( {x}^{y} ) = y \: }}$

$\boxed{\sf{  \:\rm \: log_{ {x}^{z} }( {x}^{y} ) = \frac{y}{z} \: }}$

$\boxed{\sf{  \:\rm \: log_{ {x}^{z} }( {w}^{y} ) = \frac{y}{z} log_{x}(w) \: }}$

$\boxed{\sf{  \:\rm \: {a}^{ log_{a}(x) } = x \: \: }}$

$\boxed{\sf{  \:\rm \: {a}^{y log_{a}(x) } = {x}^{y} \: \: }}$

Answer 2

$\color{orangered}\[ \begin{array}{l} \begin{array}{l} \tt \log _{2}(x-1)=1+\log _{2}(2 x-3) \\ \\ \tt \log _{2}(x-1)=\log _{2} 2+\log _{2}(2 x-3) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \qquad \qquad \qquad \qquad\left(\because \log _{u} a=1\right) \end{array} \\ \\ \tt\log _{2}(x-1)=\log _{2}[2 \times(2 x-3)] \\ \\ \tt\text { Comparing, } \\ \\ \tt x-1=2(2 x-3) \\ \\ \tt x-1=4 x-6 \\ \\ \tt\Rightarrow 4 x-x=6-1 \\ \\ \tt \Rightarrow 3 x=5 \Rightarrow x=\dfrac{5}{3} \\ \\ \tt \therefore x=\dfrac{5}{3} \end{array} \]$