Question

solve for x: 5logx+3logx=3logx+1 - 5logx-1​

Answer 1

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \mathfrak{ \: \: Given : \mapsto}} \\ \\ \mathtt{5 {}^{(log \: x)} + 3 {}^{(log \: x)} = 3 {}^{(log \: x + 1)} - 5 {}^{(log \: x- 1)} } \\ \\ \underline{ \mathfrak{ \: \: Suppose : \mapsto }} \\ \\ \mathtt{We \: \: \: assume \: \: \: that, \: \underline{ \: log \: x = a \: }}$

$\: \: \: \: \: \:\: \: \: \: \mathsf{5 {}^{(log \: x)} + 3 {}^{(log \: x)} = 3 {}^{(log \: x + 1)} - 5 {}^{(log \: x- 1)} } \\ \\ \mathsf{\Longrightarrow \: 5 \: {}^{a} + 3 \: {}^{a} = 3 {}^{( a+ 1)} - 5 {}^{(a - 1)} } \\ \\ \mathsf{\Longrightarrow \: 5 \: {}^{a} +5 {}^{(a - 1)} =3 {}^{( a+ 1)} - 3 \: {}^{a}} \\ \\ \mathsf{ \Longrightarrow \: 5 {}^{a} + 5 {}^{a}.5 {}^{( - 1)} = 3 {}^{a} .3 {}^{1} - 3 {}^{a} }\\ \\ \mathsf{\Longrightarrow \: 5 {}^{a} + 5 {}^{a}. \frac{1}{5} = 3 {}^{a}.3 - 3 {}^{a} } \\ \\ \mathsf{\Longrightarrow \: 5 {}^{a}(1 + \frac{1}{5} ) = 3 {}^{a}(3 - 1) } \\ \\ \mathsf{\Longrightarrow \: 5 {}^{a} \times \frac{6}{5} = 3 {}^{a} \times 2} \\ \\ \mathsf{ \Longrightarrow \: \frac{ \: 5 {}^{a} \: }{3 {}^{ a} } = \cancel{2} \times \frac{5}{ \cancel{6}} _3 } \\ \\ \mathsf{\Longrightarrow \: ( \frac{5}{3} ) {}^{a} = \frac{5}{3} } \\ \\ \mathsf{\Longrightarrow \: ( \frac{5}{3} ) {}^{a} =( \frac{5}{3} ){}^{1} } \\ \\ \: \: \: \: \: \: \: \: \mathsf{ \therefore \: \underline{ \: \: a = 1 \: \: }}$

$\text{ \:Now, \: } \\ \\ \text{ \underline{ \: \:We \: \: have \: assumed \: \: that, }} \\ \\ \: \: \: \: \: \: \: \: \: \: \mathtt{ log \: x = a} \\ \\ \mathtt{\Longrightarrow \: log \: x =1 } \\ \\ \mathtt{\Longrightarrow \: log \: x = log \: 10} \\ \\ \: \: \: \: \: \: \: \mathtt{ \therefore \: \: \underline{ \: \: x = 10 \: \: }}$

$\underline{ \underline{ \: : \star : \: \mathcal{ \red{PROPERTY \: \: \: USED}} \: : \star : \: }} \\ \\ \bold{1. \:If \: \: p {}^{m + n} = p {}^{m} \times p{}^{n} } \\ \\ \bold{2. \: If \: \: p {}^{m} = p {}^{n} \: \: then \: \: m = n } \\ \\ \bold{3. \: log \: 10 = 1} \\ \\ \: \bold{4. \: p \ {}^{ - 1} = \frac{1}{p} }$