Problem Solving

$$if\ \ \left(4^x\right)\left(8^y\right)\left(16^{7x}\right)=\left(8^{4y}\right)\left(32^x\right)\left(64^y\right)$$
$$what\ is\ the\ value\ of\ \ \frac{x}{y}\ ?$$
A. 3/5
B. 2/3
C. 3/4
D. 1
E. 4/3
OA is a

which of the option here is accurate and best fit? kindly help me out Thank you

PRIOR POSTS IN THIS TOPIC

Roland2rule
Exponential/power
Posted Fri Nov 03, 2017 12:16 am
$$if\ \ \left(4^x\right)\left(8^y\right)\left(16^{7x}\right)=\left(8^{4y}\right)\left(32^x\right)\left(64^y\right)$$
$$what\ is\ the\ value\ of\ \ \frac{x}{y}\ ?$$
A. 3/5
B. 2/3
C. 3/4
D. 1
E. 4/3
OA is a

which of the option here is accurate and best fit? kindly help me out Crying or Very sad Thank you

Hi Roland2rule,
Let's take a look at your question.

$$\left(4^x\right)\left(8^y\right)\left(16^{7x}\right)=\left(8^{4y}\right)\left(32^x\right)\left(64^y\right)$$
$$\left(\left(2^2\right)^x\right)\left(\left(2^3\right)^y\right)\left(\left(2^4\right)^{7x}\right)=\left(\left(2^3\right)^{4y}\right)\left(\left(2^5\right)^x\right)\left(\left(2^6\right)^y\right)$$
$$\left(2^{2x}\right)\left(2^{3y}\right)\left(2^{28x}\right)=\left(2^{12y}\right)\left(2^{5x}\right)\left(2^{6y}\right)$$
$$\left(2^{2x+3y+28x}\right)=\left(2^{12y+5x+6y}\right)$$
$$\left(2^{30x+3y}\right)=\left(2^{5x+18y}\right)$$

Since, the base is the same on both sides of the equations, we can equate the exponents, therefore,
$$30x+3y=5x+18y$$
$$30x+3y-5x=18y$$
$$25x+3y=18y$$
$$25x=18y-3y$$
$$25x=15y$$
$$\frac{x}{y}=\frac{15}{25}$$
$$\frac{x}{y}=\frac{3}{5}$$

Therefore, Option A is correct.

Hope it helps.
I am available if you'd like any follow up.