The logarithm of the logarithm of sq. root of $x$ plus $5$ plus sq. root of $x$ to the bottom $5$, to the bottom $7$ equals to zero is the given logarithmic equation on this downside.

$log_{7}{Huge(log_{5}{massive(sqrt{x+5}+sqrt{x}massive)}Huge)}$ $,=,$ $0$

The worth of the variable $x$ is unknown on this equation and it must be solved on this logarithmic equation downside. So, let’s learn to discover the worth of the variable $x$ by fixing this logarithmic equation.

Take away the Logarithms from the equation

The logarithmic kind within the equation initially creates the hurdles for fixing the equation. Subsequently, eliminating the logarithmic kind is a recommendable trick for fixing the given logarithmic equation. It may be achieved through the use of the mathematical relationship between logarithms and exponents.

$implies$ $log_{5}{massive(sqrt{x+5}+sqrt{x}massive)}$ $,=,$ $7^{0}$

Based on the zero energy rule, seven raised to the ability of zero is the same as one.

$implies$ $log_{5}{massive(sqrt{x+5}+sqrt{x}massive)}$ $,=,$ $1$

Get rid of the Logarithm utterly from equation

The logarithm remains to be concerned within the equation with base 5. So, use the relation between the logarithms and exponents another time for utterly eradicating the logarithm from the equation.

$implies$ $sqrt{x+5}$ $+$ $sqrt{x}$ $,=,$ $5^1$

Clear up the irrational equation by simplification

The given logarithmic equation is efficiently transformed into an irrational equation by releasing the equation from the logarithms.

$implies$ $sqrt{x+5}$ $+$ $sqrt{x}$ $,=,$ $5$

Take a look at the algebraic expression on the left hand facet of the equation. It’s utterly in irrational kind and this irrational kind creates downside for locating the worth of the variable $x$. So, it’s endorsed to get rid of the irrational kind from the equation.

Shift the sq. root of $x$ from the left hand facet to proper hand facet of the equation by transposition for initiating the method of eliminating the irrational kind the equation utterly.

$implies$ $sqrt{x+5}$ $,=,$ $5-sqrt{x}$

Take the sq. for the each expressions with a view to attempt to get rid of the sq. root from the equation.

$implies$ $(sqrt{x+5})^2$ $,=,$ $(5-sqrt{x})^2$

$implies$ $x+5$ $,=,$ $(5-sqrt{x})^2$

On the appropriate hand facet of the equation, increase the sq. of distinction of the phrases through the use of the sq. of distinction rule.

$implies$ $x+5$ $,=,$ $5^2$ $+$ $(sqrt{x})^2$ $-$ $2 occasions 5 occasions sqrt{x}$

Now, simplify the mathematical equation in algebraic kind by transposing.

$implies$ $x+5$ $,=,$ $25$ $+$ $x$ $-$ $10sqrt{x}$

$implies$ $10sqrt{x}$ $,=,$ $25$ $+$ $x$ $-$ $x$ $-$ $5$

$implies$ $10sqrt{x}$ $,=,$ $25$ $-$ $5$ $+$ $x$ $-$ $x$

$implies$ $10sqrt{x}$ $,=,$ $20$ $+$ $cancel{x}$ $-$ $cancel{x}$

$implies$ $10sqrt{x}$ $,=,$ $20$

$implies$ $10 occasions sqrt{x}$ $,=,$ $20$

$implies$ $sqrt{x}$ $,=,$ $dfrac{20}{10}$

$implies$ $sqrt{x}$ $,=,$ $dfrac{cancel{20}}{cancel{10}}$

$implies$ $sqrt{x}$ $,=,$ $2$

The equation is efficiently simplified by it’s nonetheless in irrational kind. So, use the identical method to get rid of the irrational kind the equation by taking sq. on either side.

$implies$ $(sqrt{x})^2$ $,=,$ $2^2$

$,,,subsequently,,,,,,$ $x$ $,=,$ $4$

LEAVE A REPLY

Please enter your comment!
Please enter your name here