Solve 10g^381-log_{e}({e})+log_{10}({1000})

$$10 { g }^{ 3 } 81- \log_{ e }({ e }) + \log_{ 10 }({ 1000 })$$

Evaluate

$$810g^{3}+2$$

Solution Steps

Multiply 10 and 81 to get 810.

$$810g^{3}-\log_{e}\left(e\right)+\log_{10}\left(1000\right)$$

The base e logarithm of e is 1.

$$810g^{3}-1+\log_{10}\left(1000\right)$$

The base 10 logarithm of 1000 is 3.

$$810g^{3}-1+3$$

Add -1 and 3 to get 2.

$$810g^{3}+2$$

Factor

$$2\left(405g^{3}+1\right)$$

Solution Steps

Multiply 10 and 81 to get 810.

$$factor(810g^{3}-\log_{e}\left(e\right)+\log_{10}\left(1000\right))$$

The base e logarithm of e is 1.

$$factor(810g^{3}-1+\log_{10}\left(1000\right))$$

The base 10 logarithm of 1000 is 3.

$$factor(810g^{3}-1+3)$$

Add -1 and 3 to get 2.

$$factor(810g^{3}+2)$$

Factor out 2. Polynomial 405g^{3}+1 is not factored since it does not have any rational roots.

$$2\left(405g^{3}+1\right)$$