What is the answer to 10^{y+1}=e^2log_{10}(e) ?

The answer to 10^{y+1}=e^2log_{10}(e) is y=(ln(e^2log_{10}(e)))/(ln(10))-1

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10^{y+1}=e^2log_{10}(e)

1 0^{y + 1} = e^{2} lo g_{10} (e)

y = \frac{ln ( e ^{2} lo g _{10} ( e ) )}{ln ( 10 )} - 1

Decimal Notation

y = - 0.49362 \dots

Solution steps

1 0^{y + 1} = e^{2} lo g_{10} (e)

Apply exponent rules

(y + 1) ln (10) = ln (e^{2} lo g_{10} (e))

Solve

(y + 1) ln (10) = ln (e^{2} lo g_{10} (e)) : y = \frac{ln ( e ^{2} lo g _{10} ( e ) )}{ln ( 10 )} - 1

y = \frac{ln ( e ^{2} lo g _{10} ( e ) )}{ln ( 10 )} - 1

6 4^{- x} = 4

2^{x + 2} + 2^{x - 3} = 66

7^{- x + 2} = 1 5^{- 2 x}

2 x e^{x} - 3 e^{x} = 0

4^{x} - 2^{x + 1} - 3 = 0

What is the answer to 10^{y+1}=e^2log_{10}(e) ?
The answer to 10^{y+1}=e^2log_{10}(e) is y=(ln(e^2log_{10}(e)))/(ln(10))-1