Can someone please explain to me how to solve logarithmic equations? Some examples are: -5=log2X
3^5x+1= 40 and log6X+(log6)7=(log6)35
I hope that makes sense because logs are hard to type.
-5=log2X
-5 = log2x
2^-5 = x
x = 1/32
3^(5x+1)= 40
let's take the log of both sides
log(3^(5x+1)) = log40
(5x + 1) log3 = log40
5x+1 = log40/log3
etc.
log6X+(log6)7=(log6)35
log6(7x) = log6 35
"antilog" both sides
7x = 35
etc.
Certainly! I'll explain how to solve each of the logarithmic equations you mentioned.
1. -5 = log2(X):
To solve this equation, you need to isolate the logarithm on one side. Start by rewriting the equation in exponential form, which can be done by raising the base (2) to the power of the logarithmic term (-5). This gives you:
2^(-5) = X
Simplifying 2^(-5) gives you the value of X:
1/32 = X
So, X = 1/32.
2. 3^(5x+1) = 40:
To solve this equation, first isolate the exponent term (5x+1). Take logarithm (with any base) of both sides to bring the exponent down. For clarity, let's choose the common logarithm with base 10 (log10):
log10(3^(5x+1)) = log10(40)
Apply the logarithmic property, which states that the logarithm of a number raised to an exponent equals the exponent multiplied by the logarithm of the number:
(5x+1) log10(3) = log10(40)
Now, divide both sides by log10(3) to solve for (5x+1):
5x + 1 = log10(40) / log10(3)
Finally, subtract 1 from both sides to obtain the value of x:
5x = (log10(40) / log10(3)) - 1
Divide both sides by 5 to solve for x:
x = [(log10(40) / log10(3)) - 1] / 5
So, x = [(log10(40) / log10(3)) - 1] / 5.
3. log6(X) + log6(7) = log6(35):
To solve this equation, you can use the logarithmic property that states the sum of logarithms of the same base is equal to the logarithm of their product.
Combine the two logarithms on the left side using this property:
log6(X * 7) = log6(35)
Simplify the equation:
log6(7X) = log6(35)
Now that the bases are the same, we can equate the arguments of the logarithms:
7X = 35
Finally, divide both sides by 7 to solve for X:
X = 35 / 7
X = 5
So, X = 5.
That should explain how to solve the logarithmic equations you mentioned!