How do I solve this logarithmic equation, which has an answer of 7? [closed]

$$10 - \log_5{20} - \log_5{25\over4}$$

The $$5$$'s are the bases and the answer to this equation is $$7$$, but I don't know how to solve it.

closed as off-topic by cmk, Ak19, mrtaurho, Javi, ShaileshJul 21 at 0:28

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• What have you tried to do? – cmk Jul 20 at 13:40
• That is not an equation. – José Carlos Santos Jul 20 at 13:41
• Welcome to MSE. This is not an equation, and you don't want to solve it. An equation has an = sign, and is an assertion that two things are the same. This is an expression, and you want to evaluate it. – saulspatz Jul 20 at 13:42
• Hint: $\log_520=\log_54+\log_55=\log_54+1$ – saulspatz Jul 20 at 13:46

$$10-\log_5(20)-\log_5(25/4)=10-(\log_5(20)+\log_5(25/4))=10-(\log_5(20\cdot25/4))$$ $$=10-\log_5(125)=10-3=7$$

In the first line I used the property that $$\log_b(x)+\log_b(y)=\log_b(xy)$$

In our case $$b=5,x=20,y=\frac{25}{4}$$. Then $$xy=125$$ so that $$\log_5(125)=3.$$

$$10-\log_520-\log_5\dfrac {25}4=10-\log_54\cdot 5-log_5\dfrac {25}4=10-(\log_55+\log_54)-(\log_525-\log_54)=10-1-\log_54-2+\log_54=10-1-2=7$$.

\begin{align*} 10-\log_5(20)-\log_5(25/4) &=10-(\log_5(20)+\log_5(25/4) ) &&-a-b=-(a+b)\\&=10-\log_5(20\cdot (25/4)) &&\log_ab+\log_ac = \log_a(b\cdot c) \\ &=10-\log_5(5^3) && 20\cdot (25/4)=(20/4)\cdot 25 = 5 \cdot 5^2 \\&=10-3\cdot\log_5(5) &&\log_a(b^c) = c\cdot\log_a(b) \\&=10-3 &&\log_a(a) = 1 \\&=7 \end{align*}

Just use your arsenal of rules. Particularly the additive and subtractive ones.

$$\log_b mn = \log_b m + \log_b n$$ and $$\log_b \frac mn = \log_b m - \log_b n$$.

And of course the definition: $$\log_b b^k = k$$. You also have $$\log_b m^k = k\log_b m$$ but oddly enough you won't need it.

$$10 - \log_5{20} - \log_5{25\over4} =$$

$$10 - \log_5 5\cdot 4 - \log_5 \frac {5^2}4 =$$

$$10 - (\log_5 5 + \log_5 4) - (\log_5 5^2 - \log_5 4) =$$

$$10 - (1+ \log_54) - (2-\log_5 4) =$$

$$10 - 1 - 2 - \log_5 4 + \log_5 4 =$$

$$10 - 1 - 2$$.

Try to think about what a $$\log$$ is.

$$5^3 = 125 \text{ means the same thing as }\log_5 125=3$$

Other examples include $$\log_3 9 = 2 \text{ because }3^2 = 9$$ and $$\log_3 81 = 4 \text{ because } 3^4 = 81$$

As an exercise, try to find $$\log_2 32 = ?$$

In general, $$\log_a b = c$$ is equivalent to $$a^c=b$$, or alternatively, $$a^{\log_a b}=b$$.

Many of the exponent rules that you're familiar with, also have logarithm versions, for example: $$\log_a BC = \log_a a^{\log_a B}a^{\log_a C} = \log_a a^{\log_a B + \log_a C} = \log_a B + \log_a C$$

and for similar reasons $$\log_a\frac{B}{C} = \log_a B - \log_a C$$

There's also a power rule for logarithms, that's related to the "power of a power" rule for exponents: $$\log_aB^C = \log_a(a^{\log_a B})^C = \log_a a^{C\log_a B}= C\log_a B$$

So, putting it all together, we have... \begin{align} 10 - \log_5 20 - \log_5\frac{25}{4} &=10 - \log_5 (4\times 5) - \log_5\frac{25}{4}\\ &= 10 - (\log_5 4 + \log_5 5) - (\log_5 25 - \log_5 4)\\ &=10 - (\log_5 4 + \log_5 5) - (\log_5 5^2 - \log_5 4)\\ &=10 - (\log_5 4 + \log_5 5) - (2\log_5 5 - \log_5 4)\\ &=10 - (\log_5 4 + 1) - (2\times 1 - \log_5 4)\\ &=10 - \log_5 4 - 1 - 2 + \log_5 4)\\ &=10 - 1 - 2 + (\log_a 4 - \log_a 4)\\ &=7 \end{align}