Question

Find the solution of \[{2^x} = 7\].

Answer 1

Hint:Whenever the right and left side of the equation do not have the same base then in that case we should take “log” both the sides. The power rule is used to multiply the two logarithms and to combine the exponents. The exponential expression should be kept by itself on one side of the equation. The logarithms of both sides of the equation should be obtained and should be solved for variable.

Complete step by step solution:
A logarithm is an exponent that is written in a special way. A logarithm with base \[10\]is a common logarithm.
The product rule states that\[{\log _b}\left( {MN} \right) = {\log _b}\left( M \right) + {\log _b}\left( N\right)\].
This property denotes that logarithm of a product is the sum of the logs of its factors. The two numbers should be multiplied with the same base then the exponents must be added.
The quotient rule states that \[{\log _b}\left( {\frac{M}{N}} \right) = {\log _b}\left( M \right) + {\log_b}\left( N \right)\]
This property denotes that the log of a quotient is the difference of the log of the dividend and the divisor.
Since, the right and left side of the equation do not have the same base then in that case we should take “log” both the sides.
We have,
\[
{2^x} = 7 \\
\Rightarrow \log \left( {{2^x}} \right) = \log \left( 7 \right) \\
\]
Using the log property we have,
\[ \Rightarrow x\log \left( 2 \right) = \log \left( 7 \right)\]
Solving for “x” now we will have,
\[
\Rightarrow x = \dfrac{{\log \left( 7 \right)}}{{\log \left( 2 \right)}} \\
\Rightarrow x \approx 2.81 \\
\]
Hence the solution of \[{2^x} = 7\] is\[ \approx 2.81\].

Note: Start by the condensing the log expressions on the left into a single logarithm using the product rule. What we want is to have a single log expression on each side of the equation. Since we want to transform the left side into a single logarithmic equation, then we should use the product rule in reverse to condense it. Always check the solved values with the original logarithmic equations.