What is log_{10}(2^x+2^{-x}) ?

The solution to log_{10}(2^x+2^{-x}) is log_{10}(4^x+1)-xlog_{10}(2)

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simplify log_{10}(2^x+2^{-x})

simplify

lo g_{10} (2^{x} + 2^{- x})

lo g_{10} (4^{x} + 1) - x lo g_{10} (2)

Solution steps

lo g_{10} (2^{x} + 2^{- x})

Simplify

2^{x} + 2^{- x} : \frac{4 ^{x} + 1}{2 ^{x}}

= lo g_{10} (\frac{4 ^{x} + 1}{2 ^{x}})

Apply log rule:

lo g_{a} (\frac{x}{y}) = lo g_{a} (x) - lo g_{a} (y)

lo g_{10} (\frac{4 ^{x} + 1}{2 ^{x}}) = lo g_{10} (4^{x} + 1) - lo g_{10} (2^{x})

= lo g_{10} (4^{x} + 1) - lo g_{10} (2^{x})

Simplify

lo g_{10} (2^{x}) : x lo g_{10} (2)

= lo g_{10} (4^{x} + 1) - x lo g_{10} (2)

(x + 2)^{2} - (x + 4)^{2}

(3 x^{2} - y^{2}) 2

\frac{( y ^{2} + 2 x y )}{x ^{2}}

[2 x^{2} + 4 x - 3] + [8 x^{2} + 2 x - 5]

5 x^{2} y - 2 x y^{2}

What is log_{10}(2^x+2^{-x}) ?
The solution to log_{10}(2^x+2^{-x}) is log_{10}(4^x+1)-xlog_{10}(2)