We read your photo as:

Express as a single logarithm:

Mathematics

ScanSolve AI Answer

Answers

2 parts
i

2+log⁡372 + \log_3 7

ii

1+log⁡32+2log⁡351 + \log_3 2 + 2 \log_3 5

Step-by-step solution

6 steps

Here are the solutions to the problems.

1. Express as a single logarithm:
log⁡3+log⁡16−log⁡4−log⁡6\log 3 + \log 16 - \log 4 - \log 6

Step 1: Use the logarithm property log⁡a+log⁡b=log⁡(a⋅b)\log a + \log b = \log (a \cdot b) to combine the positive terms and the negative terms separately.

(log⁡3+log⁡16)−(log⁡4+log⁡6)(\log 3 + \log 16) - (\log 4 + \log 6) log⁡(3×16)−log⁡(4×6)\log (3 \times 16) - \log (4 \times 6) log⁡48−log⁡24\log 48 - \log 24

Step 2: Use the logarithm property log⁡a−log⁡b=log⁡(ab)\log a - \log b = \log \left(\frac{a}{b}\right) to combine the remaining terms.

log⁡(4824)\log \left(\frac{48}{24}\right)

Step 3: Simplify the fraction.

log⁡2\log 2

The expression as a single logarithm is log⁡2\boxed{\log 2}.

2. Simplify:

Part i· 3 steps

log⁡363\log_3 63

  1. Find the prime factorization of 63.

    63=9×7=32×763 = 9 \times 7 = 3^2 \times 7
  2. Apply the logarithm properties log⁡b(xy)=log⁡bx+log⁡by\log_b (xy) = \log_b x + \log_b y and log⁡bxn=nlog⁡bx\log_b x^n = n \log_b x.

    log⁡363=log⁡3(32×7)\log_3 63 = \log_3 (3^2 \times 7) log⁡3(32)+log⁡37\log_3 (3^2) + \log_3 7 2log⁡33+log⁡372 \log_3 3 + \log_3 7
  3. Since log⁡33=1\log_3 3 = 1, substitute this value.

    2(1)+log⁡372(1) + \log_3 7 2+log⁡372 + \log_3 7

    The simplified expression is 2+log⁡37\boxed{2 + \log_3 7}.

Part ii

1+log⁡32+2log⁡351 + \log_3 2 + 2 \log_3 5

log⁡3150\log_3 150

  1. Find the prime factorization of 150.

    150=15×10=(3×5)×(2×5)=2×3×52150 = 15 \times 10 = (3 \times 5) \times (2 \times 5) = 2 \times 3 \times 5^2
  2. Apply the logarithm properties log⁡b(xyz)=log⁡bx+log⁡by+log⁡bz\log_b (xyz) = \log_b x + \log_b y + \log_b z and log⁡bxn=nlog⁡bx\log_b x^n = n \log_b x.

    log⁡3150=log⁡3(2×3×52)\log_3 150 = \log_3 (2 \times 3 \times 5^2) log⁡32+log⁡33+log⁡3(52)\log_3 2 + \log_3 3 + \log_3 (5^2) log⁡32+log⁡33+2log⁡35\log_3 2 + \log_3 3 + 2 \log_3 5
  3. Since log⁡33=1\log_3 3 = 1, substitute this value.

    log⁡32+1+2log⁡35\log_3 2 + 1 + 2 \log_3 5

    The simplified expression is 1+log⁡32+2log⁡35\boxed{1 + \log_3 2 + 2 \log_3 5}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp