How to solve exponential inequalities using logarithms

In certain situations, exponential inequalities of the form $$ b^x > c $$ can be solved by applying logarithms with an arbitrary base to both sides of the inequality.

If the base of the logarithm satisfies k>1, the direction of the inequality is preserved after taking logarithms

$$ b^x > c $$

$$ \log_k b^x > \log_k c $$

$$ x = \frac{\log_k c}{\log_k b} $$

If the base of the logarithm satisfies 0<k<1, the direction of the inequality is reversed after taking logarithms

$$ b^x > c $$

$$ \log_k b^x < \log_k c $$

$$ x < \frac{\log_k c}{\log_k b} $$

Note. This method can be applied provided that both sides of the exponential inequality are positive real numbers, since the logarithm of zero or a negative number is not defined.

A practical example

Consider the exponential inequality

$$ 4^{3+x} > 7^{2-x} $$

This inequality can be solved using logarithms because both sides are positive for the values of x under consideration.

The domain of definition of the inequality is

$$ \begin{cases} x>-3 \\ \\ x<2 \end{cases} $$

I now take the logarithm, with an arbitrary base, of both sides of the inequality.

For convenience, I choose the natural logarithm.

$$ \log 4^{3+x} > \log 7^{2-x} $$

I apply the properties of logarithms

$$ (3+x) \log 4 > (2-x) \log 7 $$

$$ 3 \log 4 + x \log 4 > 2 \log 7 - x \log 7 $$

$$ x \log 4 + x \log 7 > 2 \log 7 - 3 \log 4 $$

$$ x \cdot ( \log 4 + \log 7 ) > 2 \log 7 - 3 \log 4 $$

$$ x \cdot ( \log 4 + \log 7 ) > 2 \log 7 - 3 \log 4 $$

I divide both sides of the inequality by log 4 + log 7

Since log 4 + log 7 > 0, the direction of the inequality remains unchanged.

$$ \frac{ x \cdot ( \log 4 + \log 7 ) }{\log 4 + \log 7} > \frac{ 2 \log 7 - 3 \log 4 }{\log 4 + \log 7} $$

$$ x > \frac{ 2 \log 7 - 3 \log 4 }{ \log 4 + \log 7 } $$

Note. At this step, particular attention must be paid to the sign of the divisor. If the divisor is negative, for example log 4 - log 7, the direction of the inequality must be reversed.

If desired, the inequality can be further simplified by noting that log 4 = log 2² = 2 log 2

$$ x > \frac{ 2 \log 7 - 3 \log 2^2 }{ \log 2^2 + \log 7 } $$

$$ x > \frac{ 2 \log 7 - 3 \cdot 2 \log 2 }{ 2 \log 2 + \log 7 } $$

$$ x > \frac{ 2 \log 7 - 6 \log 2 }{ 2 \log 2 + \log 7 } $$

This expression gives the solution of the exponential inequality.

And so on.