Question

Using the graph of f(x) = log10x below, approximate the value of y in the equation 10y = 6

Answer 1

f(x) = ㏒(10x)
f(x) = ㏒₁₀(10x)

10y = 6
10 10
y = 0.6

f(x) = ㏒₁₀(10x)
0.6 = ㏒₁₀(10x)

`10^(0.6) = 10x`

`10^{(3)/(5)} = 10x`

`\sqrt[5]{10^(3)} = 10x`

`\sqrt[5]{1000} = 10x`

`\frac{\sqrt[5]{1000}}{10} = x`

`(3.98)/(10) \approx x`

`0.398 \approx x`

Answer 2

The given function is

f(x)= log 10 x

Taking base as 10

`y=\log_(10)10 +\log_(10)x`→→Using the property of, log a b= log a + log b

→y = 1 +
`\log_(10)x` as,
`\log_(10)10=1`

we have to approximate the value of y in the equation

→10 y = 6

→y=
`(6)/(10)`

`(6)/(10)`= 1 +
`\log_(10)x`

`\log_(10)x` =
`(-4)/(10)`

x=
`(10)^{(-4)/(10)}`=
`(10)^(-0.4)`=0.398

Solving graphically,

we get , x= 0.398 for , y=
`(6)/(10)`=.6