What is the answer to x^{1+log_{7}(x)}=7x ?

The answer to x^{1+log_{7}(x)}=7x is x=7,x= 1/7

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x^{1+log_{7}(x)}=7x

x^{1 + l o g_{7} (x)} = 7 x

x = 7, x = \frac{1}{7}

Decimal Notation

x = 7, x = 0.14285 \dots

Solution steps

x^{1 + l o g_{7} (x)} = 7 x

Apply log rules

(1 + lo g_{7} (x)) lo g_{7} (x) = lo g_{7} (x) + 1

Rewrite the equation with

lo g_{7} (x) = u

(1 + u) u = u + 1

Solve

(1 + u) u = u + 1 : u = 1, u = - 1

u = 1, u = - 1

Substitute back

u = lo g_{7} (x),

solve for

x

Solve

lo g_{7} (x) = 1 : x = 7

Solve

lo g_{7} (x) = - 1 : x = \frac{1}{7}

x = 7, x = \frac{1}{7}

Verify Solutions:

x = 7

True

, x = \frac{1}{7}

True

The solutions are

x = 7, x = \frac{1}{7}

lo g_{10} (33 - x) = 1 + lo g_{10} (x)

3 lo g_{x} (3) = 8 - 4 lo g_{3} (x)

lo g_{3} (lo g_{2} (x)) = - 1

3 ln (x) = ln (216)

5 = lo g_{4} (x) + lo g_{4} (x) + 1

What is the answer to x^{1+log_{7}(x)}=7x ?
The answer to x^{1+log_{7}(x)}=7x is x=7,x= 1/7
Solve for x: x^{1+log_{7}(x)}=7x
The solution is x=7,x= 1/7