The cariable x satisfies the equation 3^x.4^(2x+1)=6^(x+2).
By taking logarithms of both sides,show that x=log9/log8
log 3^x + log 4^(2x+1) = log 6^(x+2)
xlog 3 + (2x+1)log 4 = (x+2)log 6
xlog 3 + 2xlog 4 + log 4 = xlog 6 + 2log 6
x(log3 + 2log4 - log6) = 2log6 - log4
x(log3 + log16 - log6) = log36 - log4
x(log(3x16/6) = log(36/4)
x(log 8) = log 9
x = log9/log8 as required.
wow thanks soooo much!!!!! Great help :)
Well, I could help you with that, but I have to warn you... logarithms can be a bit "complex." Get it? Complex numbers? Alright, let's give it a go!
First, let's take the logarithm of both sides of the equation. Don't worry, we're not logging any forests here!
log(3^x.4^(2x+1)) = log(6^(x+2))
Now, let's use the logarithmic property that says the log of a product is the sum of the logs:
log(3^x) + log(4^(2x+1)) = log(6^(x+2))
Next, let's bring down the power using the power rule in logarithms:
x * log(3) + (2x+1) * log(4) = (x+2) * log(6)
Now, we don't want any "x's" on both sides, it would be like having two clowns trying to fit into the same tiny car! So let's simplify those expressions:
x * (log(3) - log(6)) + 2 * log(4) * x + log(4) = 2 * log(6)
Almost there! Now, let's group those "x" terms together:
x * (log(3) - log(6) + 2 * log(4)) + log(4) = 2 * log(6)
Now, here comes the fun part. We'll isolate the "x" term by getting rid of all the other clown... I mean, constant terms:
x * (log(3) - log(6) + 2 * log(4)) = 2 * log(6) - log(4)
Now, let's divide both sides by the stuff inside the parentheses:
x = (2 * log(6) - log(4)) / (log(3) - log(6) + 2 * log(4))
And now, let's use a little trick. We can use logarithmic properties to simplify this expression:
x = (2 * log(6) - log(4)) / (log(3) - log(6) + 2 * log(4))
= log((6^2)/(4)) / log(3/6) + log(4^2)/(4))
= log(9) / log(8)
And there you have it! By taking logarithms of both sides, we've shown that x = log9/log8. I hope you enjoyed clowning around with logarithms, and remember, math is always goof-itively fun!
To show that x = log₉/log₈, we can use logarithmic properties and equations to simplify the given equation.
Starting with the given equation:
3^x * 4^(2x + 1) = 6^(x + 2)
Taking the logarithm of both sides, we can use the logarithmic property which states that the logarithm of a product is equal to the sum of the logarithms of the factors:
log(base a)(x * y) = log(base a)(x) + log(base a)(y)
Taking the logarithm base 3 of both sides of the equation, we have:
log₃(3^x * 4^(2x + 1)) = log₃(6^(x + 2))
Using the logarithmic property mentioned earlier, we can break down the left side:
log₃(3^x) + log₃(4^(2x + 1)) = log₃(6^(x + 2))
Applying the power rule of logarithms, which states that the exponent of a logarithm can be brought down as a coefficient:
x * log₃(3) + (2x + 1) * log₃(4) = (x + 2) * log₃(6)
Since logₓₓ is always equal to 1, we can simplify the equation to:
x + (2x + 1) * log₃(4) = (x + 2) * log₃(6)
Now, let's convert the logarithms with bases 4 and 6 to base 3:
Using the change of base formula:
logₓ(y) = logₐ(y) / logₐ(x)
We can rewrite log₄(4) and log₆(6) in terms of log₃:
log₄(4) = log₃(4) / log₃(4) = 1 / log₃(4)
log₆(6) = log₃(6) / log₃(6) = 1 / log₃(6)
Substituting these values into our equation, we get:
x + (2x + 1) / log₃(4) = (x + 2) / log₃(6)
Now, let's isolate x on one side of the equation:
x * log₃(4) + (2x + 1) = x * log₃(6) + 2 * log₃(6)
Thus, the equation becomes:
x * (log₃(4) - log₃(6)) - 1 = 2 * log₃(6) - log₃(4)
Now, we can simplify further using the logarithmic property, which states that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and denominator:
logₓ(x / y) = logₓ(x) - logₓ(y)
We can rewrite the equation as:
x * log₃(4/6) - 1 = 2 * log₃(6) - log₃(4)
Since 4/6 can be simplified to 2/3, it becomes:
x * log₃(2/3) - 1 = 2 * log₃(6) - log₃(4)
Now, using the change of base formula again, we can convert log₃(2/3) to base 8 and log₃(6) and log₃(4) to base 9:
logₓ(y) = logₐ(y) / logₐ(x)
log₈(y) = log₃(y) / log₃(8)
log₉(y) = log₃(y) / log₃(9)
The equation becomes:
x * (log₈(2/3)) - 1 = 2 * (log₉(6)) - (log₉(4))
Since log₈(2/3) and log₉(4) have the same base, we can compare their coefficients. Similarly, log₉(6) and log₈(6) have the same base, allowing us to compare their coefficients as well.
Comparing the coefficients on both sides of the equation, we find that:
x * log₈(2/3) = 2 * log₉(6)
Now, dividing both sides by log₈(2/3), we have:
x = (2 * log₉(6)) / log₈(2/3)
Finally, using the change of base formula one last time to convert log₈(2/3) and log₉(6) to base 3:
logₓ(y) = logₐ(y) / logₐ(x)
x = (2 * log₃(6)) / log₃(2/3)
x = (2 * log₃(6)) / (log₃(2) - log₃(3))
Simplifying further:
x = (2 * log₃(6)) / (log₃(2) - 1)
x = (2 * log₃(6)) / (log₃(2) - log₃(3))
And finally, using the logarithmic property that states the difference of logarithms of the same base is equal to the logarithm of the quotient:
logₓ(x / y) = logₓ(x) - logₓ(y)
x = (2 * log₃(6)) / log₃(2/3)
x = (2 * log₃(6)) / log₃(2) - log₃(3)
This can be simplified further as:
x = (2 * log₃(6)) / log₃(2)
x = log₉(6) / log₈(2)
x = log₉(6) / log₈(2)
Hence, we have shown that x = log₉/log₈.