972^x = 8, 243^y = 16, then 3/x-4/y=?

2 Answers
Apr 10, 2018

2

Explanation:

972^x=8, 243^y=16

try to eliminate 'x'

972=8^(1/x), 243=16^(1/y)

make exponent same => 2

972 = 2^(3/x), 243=2^(4/y),

then, you can figure

972/243 = 2^(3/x-4/y),

972=2^2xx3^5, 243=3^5

2^(3/x-4/y)=4,

:.3/x-4/y = 2

Apr 10, 2018

2.

Explanation:

Recall that, a^x=y iff log_ay=x.

Accordingly, 972^x=8 rArr x=log_972 8.

By the Change of Base Rule (CBR), then, x=log_b 8/log_b 972,

where b is a new base, with b gt 0, b!=1.

Similarly, y=log_b 16/log_b 243.

:. 3/x-4/y=3*log_b 972/log_b 8-4*log_b 243/log_b 16,

=3*log_b 972/log_b 2^3-4*log_b 243/log_b 2^4,

=3*log_b 972/(3log_b 2)-4*log_b 243/(4log_b 2),

=log_b 972/log_b 2-log_b 243/log_b 2,

=(log_b 972-log_b 243)/log_b 2,

=log_b (972/243)/log_b 2,

=log_b 4/log_b 2,

=log_2 4.................[because," the CBR]",

rArr 3/x-4/y=2.