What is the answer to log_{7}(x+9)+log_{7}(49)=0 ?

The answer to log_{7}(x+9)+log_{7}(49)=0 is x=-440/49

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log_{7}(x+9)+log_{7}(49)=0

lo g_{7} (x + 9) + lo g_{7} (49) = 0

x = - \frac{440}{49}

Decimal Notation

x = - 8.97959 \dots

Solution steps

lo g_{7} (x + 9) + lo g_{7} (49) = 0

Move

lo g_{7} (49)

to the right side

lo g_{7} (x + 9) = - lo g_{7} (49)

Apply log rules

x + 9 = \frac{1}{49}

Solve

x + 9 = \frac{1}{49} : x = - \frac{440}{49}

x = - \frac{440}{49}

Verify Solutions:

x = - \frac{440}{49}

True

The solution is

x = - \frac{440}{49}

4 lo g_{x} (3) + lo g_{3} (x) + 5 = 0

ln (\frac{1}{3} x + 2) = 2

ln (- x) = ln (\frac{1}{x})

ln (2 t + 1) = \frac{9}{4}

a, ln (a) + ln (b) = c

What is the answer to log_{7}(x+9)+log_{7}(49)=0 ?
The answer to log_{7}(x+9)+log_{7}(49)=0 is x=-440/49
Solve for x: log_{7}(x+9)+log_{7}(49)=0
The solution is x=-440/49