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This is one of our many quizzes on Solving exponential equations. Click on this to access more such competitive quizzes on Differentiation.

Solving exponential equations Quiz 2

mathematics

 Find the value of ${\left( {\sqrt {{x^{ - 3}}} } \right)^5}$ 

  By using exponential function ${\left( {{x^m}} \right)^n} = {x^{mn}}$ 

${\left( {\sqrt {{x^{ - 3}}} } \right)^5} \\ \Rightarrow  {\left( {\sqrt {\dfrac{1}{{{x^3}}}} } \right)^5} \\ \Rightarrow  {\left( {\dfrac{1}{{{x^{\frac{3}{2}}}}}} \right)^5} \\ \Rightarrow  \dfrac{1}{{{x^{\frac{{15}}{2}}}}} \\  $ 

 Find the value of $x$ in given inequality ${2^{5x}} > {5^{8 - 5x}}$ 

 Take log both sides and solve it as for normal inequality.

 Let’s take the base of $10$ logarithm of both sides gives  $\log {2^{5x}} > \log {5^{8 - 5x}} \\ \Rightarrow  5x\log 2 > \left( {8 - 5x} \right)\log 5 \\ $ So, $5x\log 2 > 8\log 5 - 5x\log 5$ $ \\ $ Rearranging gives $5x\left( {\log 2 + \log 5} \right) > 8\log 5\,{\text{.since log2 + log5 = log10 = 1}} $ $ \\ $ This is equivalent to $5x > 8\log 5$ , so $x > \dfrac{8}{5}\log 5$ 

 Calculate the value of a given exponential equation ${7^x} + {7^{x + 1}} = {2^a} \times {7^x}$ ?

 By using property of exponential function ${a^x}{a^y} = {a^{x + y}}$ and ${a^x} \div {a^b} = {a^{x - y}}$ 

 Using the property: ${a^x}{a^y} = {a^{x + y}}$ hence, ${7^{x + 1}}$ can be written as  ${7^x} \times 7$ the given equation is written thus: $ \\ {7^x} + {7^{x + 1}} = {7^x} + {7^x} \times 7 \\ $ Factor out the term ${3^x}$  $ \\ \Rightarrow {7^x} + {7^{x + 1}} = {7^x} + {7^x} \times 7 = {7^x}(1 + 7) \\   \Rightarrow {7^x} + {7^{x + 1}} = {7^x}(8) \\   \Rightarrow {7^x} + {7^{x + 1}} = 8 \times {7^x} \\   \Rightarrow 8 \times {7^x} = {2^3} \times {7^x} \\   \Rightarrow {2^3} \times {7^x} = {2^a} \times {7^x} \\ $  hence, $a = 3 $ 

 Calculate the value of $x$ in the given equation by using the rules of logarithm ${\log _3}2x - {\log _3}\left( {x - 3} \right) = 1$ 

 Rising given value as a power on both sides of a similar will not change the equation and using exponential function ${a^{{{\log }_a}x}} = x$ .

${\log _3}\dfrac{{2x}}{{x - 3}} = 1 \\   \Rightarrow {3^{{{\log }_3}\dfrac{{2x}}{{x - 3}}}} = {3^1} \\   \Rightarrow \dfrac{{2x}}{{x - 3}} = 3 \\   \Rightarrow 3\left( {x - 3} \right) = 2x \\   \Rightarrow x = 3 \\  $ 

 Find the solution of given equation: $ \\{\left[ {\left( {\dfrac{{{x^{p(p - q)}}}}{{{x^{p(p + q)}}}}} \right) \div \left( {\dfrac{{{x^{p(q - p)}}}}{{{x^{q(q + p)}}}}} \right)} \right]^{p + q}} = $

 Use exponential function: ${x^a} \div {x^b} = {x^{a - b}}$ and ${x^0} = 1$ 

 $ {\left[ {\left( {\dfrac{{{x^{p(p - q)}}}}{{{x^{p(p + q)}}}}} \right) \div \left( {\dfrac{{{x^{q(q - p)}}}}{{{x^{q(q + p)}}}}} \right)} \right]^{p + q}} \\ \Rightarrow  {\left[ {\left( {\dfrac{{{x^{{p^2} - pq}}}}{{{x^{{p^2} + pq}}}}} \right) \div \left( {\dfrac{{{x^{{q^2} - pq}}}}{{{x^{{q^2} + pq}}}}} \right)} \right]^{p + q}} \\ \Rightarrow  {\left[ {\left( {{x^{{p^2} - pq - ({p^2} + pq)}}} \right) \div \left( {{x^{{q^2} - pq - \left( {{q^2} + pq} \right)}}} \right)} \right]^{p + q}} \\ \Rightarrow  {\left[ {\left( {{x^{{p^2} - pq - {p^2} - pq}}} \right) \div \left( {{x^{{q^2} - pq - {q^2} - pq}}} \right)} \right]^{p + q}} \\ \Rightarrow  {\left[ {\left( {{x^{ - 2pq}}} \right) \div \left( {{x^{ - 2pq}}} \right)} \right]^{p + q}} \\ \Rightarrow  {\left[ {{x^{ - 2pq - ( - 2pq)}}} \right]^{p + q}} \\   \Rightarrow {\left[ {{x^{ - 2pq + 2pq}}} \right]^{p + q}} = {\left( {{x^0}} \right)^{p + q}} \\   \Rightarrow {x^0} = 1 \\  $ 