[Math] How to calculate $\log(x) = 1/2\log(16) – 1/3\log(8) + 1$

algebra-precalculuslogarithms

This is my first question.

This is basic math, but what I get does not match the alternatives I have. So I was wondering if I did something wrong.

Step 1:

$\log(x) = 1/2\log(16) – 1/3\log(8) + 1$

Step 2:

$\log(x) = 1/2(1.2) – 1/3(0.9) + 1$

Step 3:

$\log(x) = 1.3$

Step 4:

According to one property of logs:

$x = 10 ** 13/10$

$x = 19.95\ldots$. This is my answer. So I should assume $20$ is the correct answer? Thanks.

My alternatives are:

a) 18

b) 20

c) 10

d) 30

e) 25

Did I do something wrong? Thanks.

Best Answer

We can write $\log 16$ as $\log 2^4 = 4 \log 2$ and $\log 8$ as $\log 2^3 = 3 \log 2$. This gives us $$\log x = \frac{1}{2} \cdot 4 \log 2 - 3 \cdot \frac{1}{3} \log 2 + 1.$$ A bit of simplifying yields $$2 \log 2 - \log 2 + 1= \log 2 + 1$$

But note that $\log 10 = 1$ so we get $$\log 2 + \log 10 = \log (2\cdot 10) = \log 20$$

So $\log x = \log 20 \iff x = 20$ by applying the exponential function to both sides.

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