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Variable Separable Method

Solve : log...

Question

Solve : ${log}_{7} (2 x - 1) + {log}_{7} (2 x - 7) = 1$ ?

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Solution

${log}_{7} (2 x - 1) + {log}_{7} (2 x - 7) = 1 \Rightarrow {log}_{7} (2 x - 1) (2 - 7) = 1 [∵ l o g a + l o g b = l o g a . b] \Rightarrow {log}_{7} ({4 x}^{2} - 14 x - 2 x + 7) = 1 \Rightarrow {log}_{7} ({4 x}^{2} - 16 x + 7) = 1 \Rightarrow {4 x}^{2} - 16 x + 7 = 7^{1} [∵ {log}_{b} a = c \Leftrightarrow a = b^{c}] \Rightarrow {4 x}^{2} - 16 x = 0 \Rightarrow 4 x (x - 4) = 0 E i t h e r, o r 4 x = 0 x - 4 = 0 \Rightarrow x = 0 \Rightarrow x = 4$

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