Question

logx =a+b;logy =a-b then log(10x/y^2)

Answer 1

log(10x/y^2)=log10x - logy^2= log10+logx-2logy=1+a+b-2(a-b)=1-a+3b

properties used here are
logxy=logx+logy
logx/y=logx-logy
logx^y=ylogx
and log10=1

hope this helps:p

Answer 2

Answer:

$log\frac{10x}{y^{2} }$=log 10+3b-a

Step-by-step explanation:

Tip

• Logarithms are a different way to write exponents in mathematics. A number with a base has a logarithm that equals another number. Exponentiation's inverse is a logarithm.

Given

logx =a+b

logy =a-b

Find

log(10x/y^2)

Solution

$\log x=a+b$,

$\log y=a-b$

$\log \frac{10 x}{y^{2}}=\log 10+\log x-2 \log y=\log 10+a+b-2(a-b)=\log 10+3 b-a$

Final Answer

$log 10+3b-a$ is the value of log$(\frac{10x}{y^{2} }$

#SPJ2