Find A so that
5log(base5)16807 − 3log(base5)7 = log (base5)A
I don't even know how to start this one...should I remove the 5 and 3 in front of the logs?
log5 16807^5 - log57^3 = log5A
log5 (16807^5/7^3) = log5A
A = 16807^5 / 7^3
= 49(16807)^4
= appr 3.91 x 10^18
oh right you just have to factor out the logbase5...thank you
You are NOT factoring it out
You are saying that if
log x = log y
then x = y
To solve for A in the equation 5log(base5)16807 − 3log(base5)7 = log(base5)A, we can start by using logarithmic properties to simplify the equation.
1. Apply the power rule of logarithms:
log(baseb)x^y = y * log(baseb)x
Using this rule, we can rewrite the equation as:
log(base5)(16807^5) - log(base5)(7^3) = log(base5)A
2. Apply the product rule of logarithms:
log(baseb)(x * y) = log(baseb)x + log(baseb)y
Applying the product rule to the left side of the equation, we get:
log(base5)(16807^5 / 7^3) = log(base5)A
3. Simplify the expression inside the logarithm:
16807^5 / 7^3 = (7^5)^5 / 7^3 = 7^25 / 7^3 = 7^22
Now we have:
log(base5)(7^22) = log(base5)A
4. Apply the logarithmic property:
log(baseb)b^x = x
Using this property, we can now rewrite the equation as:
22 = log(base5)A
5. Rewrite the equation in exponential form:
A = b^x
Rewriting the equation, we have:
A = 5^22
Therefore, A is equal to 5 raised to the power of 22, which is approximately equal to 969,532,843,372,085,749,760,000.