Question

Log6(x+7)=log6(6x-13) Log2(8+3x)=log2(3-x)+1

Answer 1

1) log6(х+7)=Log6(6х-13), х+7=6х-13, -5х=-20, х=4

Answer 2

To solve the given logarithmic equations, let's solve them one by one:

1. Log6(x+7) = log6(6x-13)

Since both logarithms have the same base, we can equate their arguments:

x + 7 = 6x - 13

Now, let's solve for x:

7 + 13 = 6x - x

20 = 5x

x = 20/5

x = 4

Therefore, the solution to the first equation is x = 4.

2. Log2(8+3x) = log2(3-x) + 1

Using the logarithmic properties, we can rewrite the equation as:

log2(8+3x) - log2(3-x) = 1

Using the quotient rule of logarithms, we can combine the logarithms:

log2((8+3x)/(3-x)) = 1

Now, let's convert the logarithmic equation into an exponential equation:

2^1 = (8+3x)/(3-x)

Simplifying further:

2 = (8+3x)/(3-x)

Cross-multiply:

2(3-x) = 8+3x

6-2x = 8+3x

Rearranging terms:

6-8 = 3x + 2x

-2 = 5x

x = -2/5

Therefore, the solution to the second equation is x = -2/5.

To summarize:

1. The solution to the equation log6(x+7) = log6(6x-13) is x = 4.
2. The solution to the equation log2(8+3x) = log2(3-x) + 1 is x = -2/5.