Question

How do I solve for x? Where I have $90 = 145 + 20\log \left( {\dfrac{{50}}{x}} \right)$.

Answer 1

Hint: In order to determine the solution of the above equation, first rewrite the equation in the form ${\log _b}x = y$ by combining terms . Now convert the expression into exponential form, and to do so use the definition of logarithm that the logarithm of the form ${\log _b}x = y$ is when converted into exponential form is equivalent to ${b^y} = x$,so compare with the given logarithm value with this form and solve the equation for the variable $x$ to get the required answer.

Complete step by step solution:
To solve the given the logarithmic function $90 = 145 + 20\log \left( {\dfrac{{50}}{x}} \right)$ in one variable $\left( x \right)$, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
Lets first rewrite the equation by transposing the terms are combining them, as
$
   \Rightarrow 90 - 145 = 20\log \left( {\dfrac{{50}}{x}} \right) \\
   \Rightarrow - 55 = 20\log \left( {\dfrac{{50}}{x}} \right) \\
   \Rightarrow 20\log \left( {\dfrac{{50}}{x}} \right) = - 55 \\
 $
Now dividing sides of the equation with $20$, we get
$
   \Rightarrow \dfrac{1}{{20}} \times 20\log \left( {\dfrac{{50}}{x}} \right) = \dfrac{1}{{20}} \times - 55 \\
   \Rightarrow \log \left( {\dfrac{{50}}{x}} \right) = - \dfrac{{11}}{4} \\
 $
Now to remove the logarithm from the above equation we have convert the above logarithm form into exponential form
Since, there is no base given, it means it is a logarithm of base $10$.
$ \Rightarrow {\log _{10}}\left( {\dfrac{{50}}{x}} \right) = - \dfrac{{11}}{4}$
Let’s convert this into its exponential form to remove ${\log _{10}}$ from the equation
As we know, Any logarithmic form ${\log _b}X = y$ when converted into equivalent exponential form results in ${b^y} = X$
So in Our question we are given ${\log _{10}}\left( {\dfrac{{50}}{x}} \right) = - \dfrac{{11}}{4}$ and if compare this with ${\log _b}x = y$ we get
$
  b = 10 \\
  y = - \dfrac{{11}}{4} \\
  X = \dfrac{{50}}{x} \\
 $
We obtain our equation as
$
   \Rightarrow {\log _{10}}\left( {\dfrac{{50}}{x}} \right) = - \dfrac{{11}}{4} \\
   \Rightarrow \dfrac{{50}}{x} = {10^{ - \dfrac{{11}}{4}}} \\
 $
Solving the above equation for variable $x$, we get
$
   \Rightarrow \dfrac{{50}}{x} = {10^{ - \dfrac{{11}}{4}}} \\
   \Rightarrow x = \dfrac{{50}}{{{{10}^{ - \dfrac{{11}}{4}}}}} \\
   \Rightarrow x = \dfrac{{50}}{{{{10}^{ - 2.7}}}} \\
 $
As we know $\dfrac{1}{{{a^{ - 1}}}} = a$
$ \Rightarrow x = 50 \times {10^{2.7}}$

Therefore, solution to equation $90 = 145 + 20\log \left( {\dfrac{{50}}{x}} \right)$ is equal to $x = 50 \times {10^{2.7}}$.

Additional Information:
1. Value of the 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.
${\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)$
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
$n\log m = \log {m^n}$
6. The above guidelines work just if the bases are the equivalent. For example, the expression ${\log _d}(m) + {\log _b}(n)$ can't be improved, on the grounds that the bases (the "d" and the "b") are not the equivalent, similarly as x2 × y3 can't be disentangled on the grounds that the bases (the x and y) are not the equivalent.

Note: 1. Don’t forgot to cross check your result
2. Remember $\log $ and $\ln $ both are different things. $\ln $ is considered as the natural logarithm having base $e$ and on the other hand $\log $ is having base $10 $ by default.
3. Logarithm of constant 1 is equal to zero.
4. Logarithm and exponent are basically the inverse of each other.