Question

Sum of all the solution(s) of the equation log10(x) + log10(x + 2) – log10(5x + 4) = 0 is-

Answer 1

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

Given Logarithmic equation is

$\rm :\longmapsto\:log_{10}(x) + log_{10}(x + 2) - log_{10}(5x + 4) = 0$

can be rewritten as

$\rm :\longmapsto\:log_{10}(x) + log_{10}(x + 2) = log_{10}(5x + 4)$

Let first define the domain of given equation :-

We know, logx is defined if x > 0

So,

$\rm :\longmapsto\:x > 0 \: \: and \: \: x + 2 > 0 \: \: and \: \: 5x + 4 > 0$

$\rm :\longmapsto\:x > 0 \: \: and \: \: x > - 2 \: \: and \: \: x > - \dfrac{4}{5}$

So,

$\bf\implies \:x > 0$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\:log_{10}\bigg[x(x + 2)\bigg] = log_{10}(5x + 4)$

We know,

$\boxed{\tt{ log_{a}(x) = log_{a}(y)\rm \implies\:x = y \: }}$

So, using this identity, we get

$\rm :\longmapsto\:x(x + 2) = 5x + 4$

$\rm :\longmapsto\: {x}^{2} + 2x= 5x + 4$

$\rm :\longmapsto\: {x}^{2} + 2x - 5x - 4 = 0$

$\rm :\longmapsto\: {x}^{2} - 3x - 4 = 0$

$\rm :\longmapsto\: {x}^{2} - 4x + x - 4 = 0$

$\rm :\longmapsto\:x(x - 4) + 1(x - 4) = 0$

$\rm :\longmapsto\:(x - 4)(x+ 1) = 0$

$\bf\implies \:x = 4 \: \: or \: \: x = - 1 \{rejected \}$

So, Sum of all the solutions = 4

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More to know

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

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

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

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

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

$\boxed{\tt{ {e}^{logx} = x \: }}$

$\boxed{\tt{ {e}^{ylogx} = {x}^{y} \: }}$