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How do you solve
$\log 2 + \log (4 x - 1)$

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130 = x

125.5 = x

x = 140

x = 150

answer is B.

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Detailed Solution

\Rightarrow \log 2 + \log (4 x - 1)

(Given) — (|)

\Rightarrow \log 2 (4 x - 1)

(log ab = log a + log b)

\Rightarrow \log 8 x - 2

— (||)
Let R.H.S be

3

We know that

\log 10 = 1

\Rightarrow 3 \log 10 = 3

(n log a = \log a^{n})

\Rightarrow 3 \log 10 = \log {(10)}^{3}

3 = \log {(10)}^{3}

— (|||)
Equating (||), (|||)

\log 8 x - 2 = \log {(10)}^{3}

8 x - 2 = 1000

8 x = 1002

x = \frac{501}{4}

∴ x = 125.25

Hence, the value of

x

for equation

\log 2 + \log (4 x - 1) = 3

\frac{501}{4}

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