Question

Suppose \(a = log5,\;b = log2\) and \(c = log\;3\) then calculate the value of 5^x = 6^x+2.
1. \(\frac{{2\left( {a + b} \right)}}{{c - a + b}}\)
2. \(\frac{{2\left( {b + c} \right)}}{{a - b - c}}\)
3. \(\frac{{2\left( {b + c} \right)}}{{c + a + b}}\)
4. \(\frac{{2\left( {b - c} \right)}}{{c + a + b}}\)
5. None of these

Answer 1

Correct Answer - Option 2 : \(\frac{{2\left( {b + c} \right)}}{{a - b - c}}\)

Concept:

Formula of Logarithm:

\({a^b} = x\; \Leftrightarrow lo{g_a}x = b\), here a ≠ 1 and a > 0 and x be any number.

Properties of Logarithms:

1. \({\log _a}a = 1\)
2. \({\log _a}\left( {x.y} \right) = {\log _a}x + {\log _a}y\)
3. \({\log _a}\left( {\frac{x}{y}} \right) = {\log _a}x - {\log _a}y\)
4. \({\log _a}\left( {\frac{1}{x}} \right) = - {\log _a}x\)
5. \({\rm{lo}}{{\rm{g}}_a}{x^p} = p{\rm{lo}}{{\rm{g}}_a}x\)
6. \(lo{g_a}\left( x \right) = \frac{{lo{g_b}\left( x \right)}}{{lo{g_b}\left( a \right)}}\)

 

Calculation:

Given: \(a = log5,\;b = log2,\;c = log\;3\)

5^x = 6^{x + 2} (Given)

Taking log on both sides we have,

\(log\;{5^x} = log\;{6^{x + 2}}\)

Using the power rule this can be written as,

\(\begin{array}{l} xlog\;5 = \left( {x + 2} \right)log6\\ xlog\;5 = \left( {x + 2} \right){\rm{log}}\left( {2 \cdot 3} \right) \end{array}\)

Applying the product rule of logarithm,

\(xlog\;5 = \left( {x + 2} \right)\left( {\log 2 + \log 3} \right)\)

By substituting \(a = log5,\;b = log2,\;c = log\;3\) in the above equation we get,

⇒ a × x = (x + 2)(b + c)

⇒ a × x = (bx + 2b + cx + 2c)

⇒ ax – bx – cx - 2(b + c) = 0

⇒ x(a – b - c) - 2(b + c) = 0

⇒ x = (2(b + c)) / (a – b - c)