Question

Solve ${\left( {x + 1} \right)^{\log \left( {x + 1} \right)}} = 100\left( {x + 1} \right)$ base is $10$.

Answer 1

Hint: In this type of question we have to use the concept of logarithm first compress the equation putting a small equation as a variable and use the value at the end. Take log both sides of the equation and solve using properties of log to obtain value for x accordingly.

Complete step-by-step answer:
ln given question
Put $y = x + 1$
${y^{\log y}} = 100y$
Can be written as
${y^{\log y - 1}} = 100$
Apply log on both sides
$\left( {\log y - 1} \right)\log y = \log 100$
Multiply log y with terms in braces we have,
$\log \left( {{y^2}} \right) - \log y = \log 100$
We can write
$
  \log 100 = \log {10^2} \\
   = 2 \\
 $
Putting value =2 we have value for log y
$\log y = \dfrac{{1 \pm \sqrt {1 + 4 \times 2} }}{2}$
Simplifying it we get,
$\log y = \dfrac{{1 \pm 3}}{2}$
Now log can have two values +or –
Finding each we have
$\log y = \dfrac{4}{2}or\dfrac{{ - 2}}{2}$
Or
$\log y = 2or - 1$
$y = {10^2}or{10^{ - 1}}$
Now put value of y we have,
$x + 1 = 100 or\dfrac{1}{{10}}$
Subtracting 1 when taken after = sign
$x = 100 - 1 = 99$
Or $x + 1 = \dfrac{1}{{10}}$
$ \Rightarrow x = \dfrac{1}{{10}} - 1 = \dfrac{{ - 9}}{{10}}$
Therefore we have,
$x = 99$ or $\dfrac{{ - 9}}{{10}}$

Additional Information: Logarithmic functions are the inverses of exponential functions. The inverse of exponential function for example $y = {a^x}$ $x = {a^y}$ the logarithmic functions are used to solve exponential equations and to explore the properties of exponential functions. They will also become extremely valuable in calculus, where they will be used to calculate the slope of certain functions and the area bounded by certain curves. It is a one to one function graph passed through a horizontal line test for functional inverse.

Note: We can use any variable in place of y just to make the equation simple and easy to solve, to obtain a value of x the square root formula is used which has two values one in negative another positive both or any one will be true for the equation.