What is 2Log5(5x^3)+1/3log5(x^2+6 written as a single logarithm?

Mathematics

2 answers:

AveGali [126]2 years ago

5 0

An important rule of logs is a*log b = log b^a.

Thus, 2 (log to the base 5 of )(5x^3) = (log to the base 5 of ) (5x^3)^2, or

(log to the base 5 of ) (25x^6).

Next, (1/3) (log to the base 5 of ) (x^2+6) = (log to the base 5 of ) (x^2+6)^(1/3).

Here, the addition in the middle of the given expression indicates multiplication:

2Log5(5x^3)+1/3log5(x^2+6) = (log to the base 5 of ) { (5x^3)^2 * (x^2+6)^(1/3) }.

Here we've expressed the given log quantity as a single log.

Vikentia [17]2 years ago

4 0

Answer: B on Edge. log5 (25x^6) 3 sqrt x^2+8. Hope this helps!

You might be interested in

(sin teta + sec teta)^ + (cos teta+ cosec teta )^ = (1 + sec x cosec)^

ArbitrLikvidat [17]

$(sin\theta+sec\theta)^2+(cos\theta+cosec\theta)^2=(1+sec\theta\ cosec\theta)^2\\\\L=sin^2\theta+2sin\theta\cdot\frac{1}{cos\thewta}+\frac{1}{cos^2\theta}+cos^2\theta+2cos\theta\cdot\frac{1}{sin\theta}+\frac{1}{sin^2\theta}\\\\=(sin^2\theta+cos^2\theta)+\frac{2sin\theta}{cos\theta}+\frac{2cos\theta}{sin\theta}+\frac{1}{cos^2\theta}+\frac{1}{sin^2\theta}$

$=1+\frac{2sin^2\theta+2cos^2\theta}{sin\theta\ cos\theta}+\frac{sin^2\theta+cos^2\theta}{sin^2\theta\ cos^2\theta}\\\\=1+\frac{2(sin^2\theta+cos^2\theta)}{sin\theta\ cos\theta}+\frac{1}{sin^2\theta\ cos^2\theta}\\\\=1+\frac{2}{sin\theta\ cos\theta}+\frac{1}{sin^2\theta\ cos^2\theta}\\\\=1^2+2sec\theta\ cosec\theta+sec^2\theta\ cosec^2\theta\\\\=(1+sec\theta\ cosec\theta)^2=R$

4 0

3 years ago

If you need more than 60% of the students surveyed to like your product how many out of 500 would you have to survey

Mashutka [201]

Here is your answer.