Question

pls answer this question ​

Answer 1

$\underline{\text{It can be solved in two ways.}}$

$\boxed{\bold{1.}}$

$\mathrm{Now,\:log_{\sqrt{5}}x=4}$

$\to \mathrm{(\sqrt{5})^{(log_{\sqrt{5}}x)}=(\sqrt{5})^{4}}$

$\to \mathrm{x = (5^{\frac{1}{2}})^{4}}$

$\to \mathrm{x = 5^{\frac{4}{2}}}$

$\to \mathrm{x = 5^{2}}$

$\to \mathrm{x = 25}$

$\boxed{\bold{2.}}$

$\mathrm{Now,\:log_{\sqrt{5}}x=4}$

$\to \mathrm{\frac{log_{e}x}{log_{e}\sqrt{5}}=4}$

$\to \mathrm{log_{e}x=4*log_{e}5^{\frac{1}{2}}}$

$\to \mathrm{log_{e}x=log_{e}5^{\frac{4}{2}}}$

$\to \mathrm{log_{e}x=log_{e}5^{2}}$

$\to \mathrm{log_{e}x=log_{e}25}$

$\to \mathrm{x=25}$

$\underline{\text{Formulas :}}$

$\mathrm{1.\:log_{a}b=\frac{log_{m}b}{log_{m}a}}$

$\mathrm{2.\:a^{log_{a}b}=b}$

Answer 2

${\bold{\green{\huge{Your\:Answer}}}}$

25

step-by-step explanation:

It is simple logarithmic equation.

we know that,

in a logarithmic quantity,

for example,

$log_ab$

b > 0 , a ≠ 1 and a >0

now,

considering the given equation,

$log_{root5}X$ = 4

we know that,

if,

$log_am$ = b

then,

m = ${a}^{b}$

so,

according to this,

X = ${(root5)}^{4}$

=> X = ${5}^{4/2}$

=> X = ${5}^{2}$

=> X = 25