Question

Log

Answer 1

twig

Answer 2

calc

Answer 3

\[\huge 5^{logx}-3^{\log-1} = 3^{logx+1} - 5^{logx-1}\]

Answer 4

Bring 3's and 5's together that would the start

Answer 5

@mathslover

Answer 6

log-1 ?

is it log (-1) ?

Answer 7

logx -1

Answer 8

Okay..

Answer 9

Seems easy now...

Let logx = y

\(5^{y} -3^{y-1} = 3^{y+1} -5^{y-1} \)

Answer 10

\[\huge 5^{logx} -3^{logx-1} = 3^{logx+1} - 5^{logx-1}\]

Answer 11

wish i thought of that sooner @mathslover

Answer 12

It ,Didn't strike me before

Answer 13

\(\large 5^{y} + 5^{y-1} = 3^{y+1 } + 3^{y-1} \)

Answer 14

\(5^{y} \left( 1 + \cfrac{1}{5} \right) = 3^{y} \left( 3 + \cfrac{1}{3} \right) \)

\(\left(\cfrac{5}{3} \right)^ y =\cfrac{10}{3} \times \cfrac{5}{6} \)

Answer 15

Simplify RHS
Equate the powers and find y
Equate y = log x
Find x

Answer 16

\[\huge [\frac{ 5 }{ 3 }]^{y} = \frac{ 25 }{ 9 }\]

Answer 17

@mathslover are u there

Answer 18

Sorry, was away!

Answer 19

\(\left(\cfrac{5}{3}\right)^y = \left(\cfrac{5}{3}\right)^2 \)

Thus, y = 2

\(\log x = 2\)

Answer 20

so 100

Answer 21

tysm

Answer 22

Yeah, if the base is 10, then it will be \(10^2\) . If the base is "e" then it will be \(e^{2} \)

Answer 23

You're welcome. Good Luck! @No.name