Question

How do you solve ${5^{x + 2}} = {4^{x + 1}}$?

Answer 1

Hint:In order to solve the above exponential function, we will first take natural logarithm on both side of the equation ,after taking logarithm use the property of logarithm which states that$\ln {m^n} = n\ln m$,so rewriting equation and using distributive law to separate out the term as $a(b + c) = ab + ac$.Now transposing the term having x on the left hand side and constants terms on the right hand side of the equation, combine like terms and transposing everything from LHS to RHS except variable x will give your required solution

Complete step by step solution:
We are Given an exponential function${5^{x + 2}} = {4^{x + 1}}$ ,
${5^{x + 2}} = {4^{x + 1}}$

Now taking natural logarithm on the both sides, we get
$\ln {5^{x + 2}} = \ln {4^{x + 1}}$

Now using property of logarithm on both side$\ln {m^n} = n\ln m$
$\left( {x + 2} \right)\ln 5 = \left( {x + 1} \right)\ln 4$

Using property of distributive $a(b + c) = ab + ac$
$x\ln 5 + 2\ln 5 = x\ln 4 + \ln 4$

Now transposing terms having x on the left-hand side and constants term on the right hand side
\[
x\ln 5 - x\ln 4 = \ln 4 - 2\ln 5 \\

\dfrac{{x(\ln 5 - \ln 4)}}{{(\ln 5 - \ln 4)}} = \dfrac{{\ln 4 - 2\ln 5}}{{(\ln 5 - \ln 4)}} \\
x = \dfrac{{(\ln 4 - 2\ln 5)}}{{(\ln 5 - \ln 4)}} \\
\]
Therefore, the solution to the exponential function${5^{x + 2}} = {4^{x + 1}}$is equal to
\[\dfrac{{(\ln 4 - 2\ln 5)}}{{(\ln 5 - \ln 4)}}\].

Additional Information: 1.Value of constant ‘e’ is equal to $2.71828$.

2.A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number, we actually undo an exponentiation.

3.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values.

${\log _b}(mn) = {\log _b}(m) + {\log _b}(n)$

4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values.

${\log _b}\left( {\dfrac{m}{n}} \right) = {\log _b}(m) - {\log _b}(n)$

5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.

$n\log m = \log {m^n}$

Note: 1.Don’t forget to cross-check your answer at least once.

2.$\ln $ is known as the “natural log” which is having base $e$