Please help simplifying logs ?

Question

a. 3^(2log[3](5)

b.log[10](10^(1/2))

c.log[10](1/10^x)

d.2log[10](√x) + 3log[10](x^(1/3))

Answer 1

a. raising to a power and logarithms are inverses of each other and since 2log...etc can be written as log[3](5)^2, the answer is 5^2 or 25.

b. log[10](10) cancel out due to inverse again so it's just 1/2

c. Well it can be rewritten as log[10](1) - log[10]10^x

the latter can be canceled to be log[10](1) - x

d. move your coefficients over so that its now log[10] (√x)^2 + log[10] (x^(1/9))

This simplifies to log[10] (x) + log[10] (x^1/9)

Then multiply out because of one of the log properties

log[10] x*x^1/9

Then that equals log[10] (x^2/9)   for your final answer.

Answer 2

Question a)

log is log to base 3

y = 3^(2 log 5)

log y = 2 log 5

log y = log 5²

y = 5²

y = 25

Question b)

log is log to base 10

y = log 10^(1/2)

y = (1/2) log 10

y = 1/2

Question c)

log is log to base 10

y = log (1 / 10^x)

y = log ( 10^(-x) )

y = - x log 10

y = - x

Question d)

log is log to base 10

y = 2 log x^(1/2) + 3 log x^(1/3)

y = log x + log x

y = 2 log x

y = log x²

Answer 3

a. 3^(2log[3](5))=3^(log[3](5²))=5²=25

b. log[10](10^(1/2))=1/2

c. log[10](1/10^x)=-log[10](10^x)=-x

d. 2log[10](√x)+3log[10](x^(1/3))

=2.1/2log[10](x)+3.1/3log[10](x)

=log[10](x)+log[10](x)

=2log[10](x)

Answer 4

a. 2log[3](5) = 10log [3] = log[3^10]

3^(log[3^10])

b. log[10] = 1

    1(10^(1/2)) = sqr. root of 10

c. log[10] = 1. 1 (1/10^x) = (1/10^x)

d. 2log[10] = 2. 3log[10] = 3.

     2(x^(1/2)) + 3(x^(1/3))