Question
Solve for $\mathbf{x}, \log _x^{15 \sqrt{5 } }=2-\log _x^{3 \sqrt{ 5} }$

Answer

$\log_x 15\sqrt 5 = 2 - \log_x3\sqrt 5$
$\Rightarrow \log_x15\sqrt 5 + \log_x3\sqrt 5 = 2$
$\Rightarrow \log_x( 15\sqrt 5 \times 3\sqrt 5 ) = 2$
$\Rightarrow \log_x225 = 2$
$\Rightarrow \log_x15^2 = 2$
$\Rightarrow 2 \log_x15= 2$
$\Rightarrow \log_x15 = 1$
$\Rightarrow x = 15.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Find the amount paid by a customer when he buys a watch priced at $Rs. 5400$ available at a discount of $12\%.$
Each wheel of a car is of diameter $80 \ cm.$ How many completer revolutions does each wheel make in $10$ minutes when the car is traveling at a speed of $66 \ km$ per hour?
Mr. Sharma borrowed a certain sum of money at $10\%$ per annum compounded annually. If by paying $Rs.19,360$ at the end of the second year and $Rs. 31,944$ at the end of the third year he clears the debt; find the sum borrowed by him.
In the adjoining figure, P is a point of intersection of two circles with centres $C$ and $D$. If the straight line APB is parallel to CD, prove that $AB =2 CD$.
Image
Solve for $\mathbf{x}: 4^{x-1} \times(0.5)^{3-2 x}=\left(\frac{1}{8}\right)^{-x}$
In History project, marks out of $20$ were awarded to $8$ students. The marks were as shown below:$14, 16, 18, 14, 16, 14, 12, 16:$Find the mean marks.
Equilateral triangles ABD and ACE are drawn on the sides $A B$ and $A C$ of $\triangle A B C$ as shown in the figure. Prove that :
(i) $\angle DAC =\angle EAB$
(ii) $DC = BE$.
Image
Simple interest on a sum of money for $2$ years at $4\%$ is $Rs. 450.$ Find compound interest of the same sum and at the same rate for $2$ years.
Solve for ' $\theta$ ': $\sin \frac{\theta}{3}=1$
The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. Also, find the cost of cultivating the field at ₹ 24.60 per $100 cm^2$.