how to solve log7y=9logy7 ?

Question

please show me the method of solving this probelm in details.

Answer 1

log[base 7](y) = 9 log[base y](7)

First off, by a log property, log[base y](7) = 1/log[base 7](y), so

log[base 7](y) = 9/log[base 7](y)

Multiply both sides by log[base 7](y) to get

{ log[base 7](y) }^2 = 9

Take the square root of both sides, remembering to include "plus or minus" when doing so.

log[base 7](y) = +/- sqrt(9)

But sqrt(9) = 3, so

log[base 7](y) = +/- 3

This gives us two equations:

log[base 7](y) = 3
log[base 7](y) = -3

Change both of them to exponential form; a reminder that
log[base b](a) = c in exponential form is b^c = a.

7^3 = y
7^(-3) = y

Therefore,

y = { 7^3, 7^(-3) }, or

y = {343, 1/343}

Answer 2

logs will be assumed as log base 7:-
log y = 9 log 7
log y = log (7^(9) )
y = 7^(9) = 4.04 x 10^(7) in standard form.

Answer 3

log7y= 9logy7
(log y)7= (9x7) log y
example: logy = p
(p)7= 63p
p7 - 63p=0
p(p6 - 63)
pi = 0 so log y =0
y = 1
pii = (63)1/6
i hope you understand what i mean,,,

Answer 4

y=any number
Please give me best answer thanks!