Question

How do you solve ${4^{3 - 2x}} = {5^{ - x}}$?

Answer 1

Hint: First take log on both sides and apply the property of log, $\log {a^b} = b\log a$. After that move all variables on one side and the constant part on another side. Then do simplification. After that divide both sides by the coefficients of $x$ to get the desired result.

Complete step-by-step solution:
Let us understand the definition of log first.
Logarithms are the opposite of exponentials, just as the opposite of addition is subtraction and the opposite of multiplication is division.
In other words, a logarithm is essentially an exponent that is written in a particular manner.
Logarithms can make multiplication and division of large numbers easier, because adding logarithms is the same as multiplying, and subtracting logarithms is the same as dividing.
The given expression is ${4^{3 - 2x}} = {5^{ - x}}$.
Take a log on both sides of the expression.
$ \Rightarrow \log {4^{3 - 2x}} = \log {5^{ - x}}$
We know that the power law of log is,
$\log {a^b} = a\log b$
Using the above law, the expression will be,
$ \Rightarrow \left( {3 - 2x} \right)\log 4 = - x\log 5$
Open the bracket and expand the terms,
$ \Rightarrow 3\log 4 - 2x\log 4 = - x\log 5$
Move the variable terms on one side,
$ \Rightarrow 2x\log 4 - x\log 5 = 3\log 4$
Take $x$ common from the left side of the expression,
$ \Rightarrow x\left( {2\log 4 - \log 5} \right) = 3\log 4$
Now, divide both sides by $\left( {2\log 4 - \log 5} \right)$ to get the value of $x$,
$\therefore x = \dfrac{{3\log 4}}{{2\log 4 - \log 5}}$

Hence, the value of x is $\dfrac{{3\log 4}}{{2\log 4 - \log 5}}$.

Note: A logarithm with base 10 is a common logarithm. In our number system, there are ten bases and ten digits from 0-9, here the place value is determined by groups of ten. You can remember common logarithms with the one whose base is common as 10.
Change of base rule law,
${\log _y}x = \dfrac{{\log x}}{{\log y}}$
Product rule law,
$\log xy = \log x + \log y$
Quotient rule law,
$\log \dfrac{x}{y} = \log x - \log y$
Power rule law,
$\log {x^y} = y\log x$