What is the answer to x^{log_{10}(4x)}-17=239 ?

The answer to x^{log_{10}(4x)}-17=239 is

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x^{log_{10}(4x)}-17=239

x^{l o g_{10} (4 x)} - 17 = 239

x = 1 0^{\frac{- 2 l o g _{10} ( 2 ) + 4 l o g _{10} ( 2 ) ^{2} + \frac{32 l n ( 2 )}{l n ( 10 )}}{2}}, x = 1 0^{\frac{- 2 l o g _{10} ( 2 ) - 4 l o g _{10} ( 2 ) ^{2} + \frac{32 l n ( 2 )}{l n ( 10 )}}{2}}

Solution steps

x^{l o g_{10} (4 x)} - 17 = 239

Apply exponent rules

lo g_{10} (4 x) ln (x) = ln (256)

Solve

lo g_{10} (4 x) ln (x) = ln (256) : x = 1 0^{\frac{- 2 l o g _{10} ( 2 ) + 4 l o g _{10} ( 2 ) ^{2} + \frac{32 l n ( 2 )}{l n ( 10 )}}{2}}, x = 1 0^{\frac{- 2 l o g _{10} ( 2 ) - 4 l o g _{10} ( 2 ) ^{2} + \frac{32 l n ( 2 )}{l n ( 10 )}}{2}}

x = 1 0^{\frac{- 2 l o g _{10} ( 2 ) + 4 l o g _{10} ( 2 ) ^{2} + \frac{32 l n ( 2 )}{l n ( 10 )}}{2}}, x = 1 0^{\frac{- 2 l o g _{10} ( 2 ) - 4 l o g _{10} ( 2 ) ^{2} + \frac{32 l n ( 2 )}{l n ( 10 )}}{2}}

Verify Solutions:

x = 1 0^{\frac{- 2 l o g _{10} ( 2 ) + 4 l o g _{10} ( 2 ) ^{2} + \frac{32 l n ( 2 )}{l n ( 10 )}}{2}}

True

, x = 1 0^{\frac{- 2 l o g _{10} ( 2 ) - 4 l o g _{10} ( 2 ) ^{2} + \frac{32 l n ( 2 )}{l n ( 10 )}}{2}}

True

The solutions are

x = 1 0^{\frac{- 2 l o g _{10} ( 2 ) + 4 l o g _{10} ( 2 ) ^{2} + \frac{32 l n ( 2 )}{l n ( 10 )}}{2}}, x = 1 0^{\frac{- 2 l o g _{10} ( 2 ) - 4 l o g _{10} ( 2 ) ^{2} + \frac{32 l n ( 2 )}{l n ( 10 )}}{2}}

x e^{x} - 2 = 0

6^{3 x + 1} = 2^{2 x - 3}

10 0^{x} = 1

3^{(x - 4)} = 7

(\frac{1}{4})^{x + 1} = 32

What is the answer to x^{log_{10}(4x)}-17=239 ?
The answer to x^{log_{10}(4x)}-17=239 is
Solve for x: x^{log_{10}(4x)}-17=239
The solution is