Question

1 point

Use log10 3-0.4771; log10 5 0.699010810 7 0.8451; log10 11 1.0414 to approximate the value of each expression-
log10 14710910 (147)

Answer 1

`\log_(10)(147) = 2.1673`

Explanation:

Given

`\log_(10) 3 = 0.4771`

`\log_(10) 5 = 0.6990`

`\log_(10) 7= 0.8451`

`\log_(10) 11 = 1.0414`

Required

Evaluate
`\log_(10)(147)`

Expand

`\log_(10)(147) = \log_(10)(49 * 3)`

Further expand

`\log_(10)(147) = \log_(10)(7 * 7 * 3)`

Apply product rule of logarithm

`\log_(10)(147) = \log_(10)(7) + \log_(10)(7) + \log_(10)(3)`

Substitute values for log(7) and log(3)

`\log_(10)(147) = 0.8451 + 0.8451 + 0.4771`

`\log_(10)(147) = 2.1673`