What is 1/(ln(a)x)(log_{10}(x)+log_{10}(x-7)-1) ?

The solution to 1/(ln(a)x)(log_{10}(x)+log_{10}(x-7)-1) is (log_{10}(x(x-7))-1)/(xln(a))

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simplify 1/(ln(a)x)(log_{10}(x)+log_{10}(x-7)-1)

simplify

\frac{1}{ln ( a ) x} (lo g_{10} (x) + lo g_{10} (x - 7) - 1)

\frac{lo g _{10} ( x ( x - 7 ) ) - 1}{x ln ( a )}

Solution steps

\frac{1}{ln ( a ) x} (lo g_{10} (x) + lo g_{10} (x - 7) - 1)

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{10} (x) + lo g_{10} (x - 7) = lo g_{10} (x (x - 7))

= \frac{1}{ln ( a ) x} (lo g_{10} (x (x - 7)) - 1)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ( lo g _{10} ( x ( x - 7 ) ) - 1 )}{ln ( a ) x}

1 \cdot (lo g_{10} (x (x - 7)) - 1) = lo g_{10} (x (x - 7)) - 1

= \frac{lo g _{10} ( x ( x - 7 ) ) - 1}{x ln ( a )}

lo g_{e} (2 π \frac{B ^{3}}{A e})

ln^{2} (\frac{1}{3}) + (ln (\frac{1}{3}))^{2}

ln (\frac{e ^{15 r} - 1}{r})

\frac{- 2 ln ( 6 )}{2 ln ( 3 ) - ln ( 6 )}

ln (\frac{33.6 \cdot 1 0 ^{6}}{21660})

What is 1/(ln(a)x)(log_{10}(x)+log_{10}(x-7)-1) ?
The solution to 1/(ln(a)x)(log_{10}(x)+log_{10}(x-7)-1) is (log_{10}(x(x-7))-1)/(xln(a))