LOGARITHMS

CONTENT

1. Revision of the Laws of Indices
2. Indices Involving Powers in Simultaneous Lnear Equations
3. Laws of Logarithms
4. Logarithmic Equations

Revision of the Laws of Indices

The expression \(a^n\) means \(a\) is multiplied by itself \(n\) times, e.g.

\(2^4 = 2 × 2 × 2 × 2 = 16\), \(3^5 = 3 × 3 × 3 × 3 ×3 = 243\)

(i) \(a^m × a^n = a^{m+n}\)   Example: \(2^5 × 2^2 = 2^{5+2} = 2^7\)

(ii) \(\frac{a^m}{a^n}\)  or 

\(a^m÷a^n=a^{m-n}\)   Example: \(\frac{3^5}{3^3}=3^{5 -3} = 3^2\)

(iii) \((a^m)^n=a^{mn} \)   Example: \((2^2)^3=2^{2×3}=2^6 \)

(iv) \(\frac{1}{a^n} =a^{-n}\)   Example: \(\frac{1}{3^2} =3^{-2}\)

(v) \(a^o = 1\)   Example: \(5^o = 1\)

(vi) \(a^{\frac{1}{n}} = \sqrt[n]{a}\)   Example: \(8^{\frac{1}{3}} = \sqrt[3]{8} = 2\)

(vii) If \(a^x = a^y\), then \(x = y\)   Example: If \(2^m = 2^n\), then \(m = n\)

Examples:

1. Evaluate \(9^{\frac{-1}{2}}\)

Solution: 

\(9^{\frac{-1}{2}}=(3^2 )^{\frac{-1}{2}}=3^{2 ×\frac{-1}{2}}=3^{-1}= \frac{1}{3}\)

2. Evaluate \(81^{\frac{1}{4}}\)

Solution:

\(81^{\frac{1}{4}} = (3^4)^{\frac{1}{4}} = 3^1 = 3\)

3. Evaluate \(\Big(\frac{1}{4}\Big)^{\frac{-1}{2}}\)

Solution:

\(\Big(\frac{1}{4}\Big)^{\frac{-1}{2}} = \Big(\frac{1}{2^2}\Big)^{\frac{-1}{2}} = (2^{-2})^{\frac{-1}{2}} = 2^{-2 × \frac{-1}{2}} = 2\)

4. Simplify \(\frac{16}{8^{\frac{-2}{3}}}\)

Solution:

\(\frac{16}{8^{\frac{-2}{3}}} = 16 ÷ 8^{\frac{-2}{3}} = 16 ÷ \frac{1}{8^{\frac{2}{3}}} \\ = 16 × 8^{\frac{2}{3}} = 2^4 × 2^{\Big(3×\frac{2}{3}\Big)} = 2^4 × 2^2 = 2^6 = 64\)

5. \(\frac{27^{\frac{2}{3}} × 81^{\frac{1}{4}}}{9^{\frac{3}{2}}}\)

Solution:

\(\frac{27^{\frac{2}{3}} × 81^{\frac{1}{4}}}{9^{\frac{3}{2}}} = \frac{(3^3)^{\frac{2}{3}} × (3^4)^{\frac{1}{4}}}{(3^2)^{\frac{3}{2}}} \\ = \frac{3^2 × 3^1}{3^3} \\ = \frac{3^3}{3^3} = 1\)

6. Simplify \((2a^2 b)^2 (-a^2 b)^3\)

Solution:

\((2a^2 b)^2 (-a^2 b)^3 = (4a^4b^2)(-a^6b^3) \\ = -4a^{4+6}b^{2+3} \\ = -4a^{10}b^5\)

7. Simplify \((2xy)^2 + (-y)(4xy)(-2x)\)

Solution:

\((2xy)^2+(-y)(4xy)(-2x) = 4x^2 y^2+(-y)(-8x^2y) \\ = 4x^2 y^2+8x^2 y^2 \\ = 12x^2 y^2\)

CLASS ACTIVITY

Simplify the following:

1. \(\sqrt[3]{\Big(\frac{8}{27}\Big)^2}\)

2. \(\sqrt[4]{\Big(5\frac{1}{16}\Big)^{-3}}\)

3. \(\frac{15a^3 b^5}{5ab^2}\)

Further Examples

Solve for \(x\):

\(3^{2x+1} = 243\) (WAEC SSCE)

Solution:

\(3^{2x+1} = 243\)

Express the R.H.S in power to base 3

\(3^{2x+1} = 3^5\)

Equating the powers, we have

\(2x + 1 = 5 \\ 2x = 5 -1 \\ 2x = 4 \\ x = 2\)

2. Solve for \(x\): \(2^{x + 2} = 8^{x -2}\) (WAEC SSCE)

Solution:

Express the R.H.S in power to base 2

\(2^{x + 2} = 2^{3(x -2)} \\ 2^{x + 2} = 2^{3x -6}\)

Equating the powers, we have

\(x + 2 = 3x -6 \\ x -3x = -6 -2 \\ -2x = -8 \\ x = 4\)

3. \(x^4 = (0.25)^2\); find \(x\).

Solution:

\(x^4 = (0.25)^2 \\ x^4 = \Big(\frac{25}{100}\Big)^2\)

We will seek to make R.H.S in powers of 4

\( x^4 = \Big(\frac{5^2}{10^2}\Big)^2 \\ x^4 = \Big(\frac{5}{10}\Big)^4 \)

Equating terms having equal powers, we have

\( x = \frac{5}{10} \\ x = 0.5\)

CLASS ACTIVITY

1. If \(4^x = 2^{\frac{1}{2}} × 8\), find \(x\)

2. If \(8^{\frac{x}{2}} = 2^{\frac{3}{8}} × 4^{\frac{3}{4}}\), find \(x\)

3. Given that \(125^{2x+1} = 625 × 25^{-x}\), find \(x\)

4. If \(3^m × 27^{(2m -1)} = 81\), find \(m\)

Indices Involving Powers in Simultaneous Lnear Equations

Example1: If \(3^{y+x} = 9^{y+x}\) and \(2^{x-y} = 8^{x -3}\), find the values of \(x\) and \(y\) respectively.

Solution:

If \(3^{y+x}=9^{y+x} \) and \(2^{x -y}=8^{x -3}\), then expressing both equations in the powers of their common base, we have

\(3{y+x}=3^{2(y+x)} ⇒ 3^{y+x}=3^{2y+2x}\) ………(i)

\(2^{x -y}=2^{3(x -3)} ⇒ 2^{x -y}=2^{3x -9}\) ……….(ii)

Equating the powers of equations (i) & (ii)

\(y+x=2y+2x ⇒x+y=0\) ………….(iii)

\(x -y=3x-9 ⇒2x+y=9\) …………(iv)

Solving equations (iii) & (iv) simultaneously, We obtain \(x = 9 , y= -9\)

Examples:

Solve the equation \(4^{x+1} -9(2^x) = -2\) (WAEC SSCE)

Solution:

\(4^{x+1} -9(2^x) = -2 \\ 4^{x+1} -9 (2^x)+2=0 \\ 4(4^x ) -9 (2^x)+2=0 \\4(2^2x) -9(2^x)+2=0 \\ 4(2^x )^2 -9(2^x)+2=0\)

Since \(2^{3×2} =(2^3 )^2 \)

let \(p=2^x\), then the equation becomes

\(4p^2 -9p+2=0\)

Factorizing we have;

\(4p^2 -8p -p+2=0 \\ 4p(p -2) -1(p -2)=0 \\ (p -2)(4p -1)=0 \\ ⇒p -2=0 \text{ or } 4p -1=0 p=2 \text{ or } p=\frac{1}{4}\)

Remember that \(p=2^x\) When \(p=2\), note that \(2\) is the same as \(2^1\).

\(2^x=2^1 \\ ∴ x = 1\)

When \(p= \frac{1}{4} \\ 2^x=\frac{1}{4} 2^x=2^{-2} ∴ x= -2\)

So, the solution of the equation is \(x=1\) and \(x= -2\)

CLASS ACTIVITY

1. Find the value of \(x\) given that \(7×49^{(x+2)} = \frac{1}{343} \)
2. Solve simultaneously for x and y; \(4^x×8^y=256\) and \(8^x×\frac{1}{4^y} =\frac{1}{2}\)
3. Find the values of \(x\) & \(y\), given that \(3^x × \frac{1}{3^y} = 9^3\) and \(9^x×3^y=27\)
4. Find the values of \(x\) & \(y\), given that \(5^x × \frac{1}{5^y} =5\) and \(5^x×5^y=125\)
5. Given that \(\frac{1}{3}\) of \(27^x=9^{2x}\), find \(x\)
6. For what values of \(x\) is \(125^{\frac{-2}{x}}=\frac{1}{25} \)?
7. Simplify \(125^{\frac{-1}{3}}×49^{\frac{-1}{2}}×10^0\)
8. Simplify \(\frac{27^{\frac{2}{3}} × 64^{\frac{-1}{2}}}{32^{\frac{-2}{5}}}\)
9. Simplify the following: (i) \(27^{\frac{1}{3}}\) (ii) \(16^{-\frac{1}{4}}\) (iii) \(125^{-\frac{1}{3}}\) (iv) \((8\frac{1}{3})^{-2}\) (v) \(y^{\frac{1}{2}} × y^{-1} × y^{\frac{3}{2}}\) (vi) Find \(x\), given that \(9×3^{1+x} = 27^{-x}\)
10. Simplify the following: (a) \(25^{\frac{-3}{2}}\) (b) \(81^{\frac{3}{4}}\) (c) \(\Big(\frac{1}{4}\Big)^{\frac{-3}{2}}\) (d) \((x^3)^{\frac{-2}{3}}\) (e) \((z^0)^{\frac{1}{4}}\) (f) \(\frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}}\) \(64^{\frac{1}{2}}×9^{\frac{3}{2}}×2^{-5}\)
11. Without using mathematical tables, simplify leaving your answer in standard form (a) \(\sqrt{\frac{8.1 × 10^{-5}}{1.44 × 10^4}}\) (b) \(\sqrt{\frac{250 × 10^7}{1.21 × 10^4}}\)
12. Simplify \(\frac{1}{3^{5n}} × 9^{n -1} × 27^{n + 1}\)
13. Solve the equation for \(x\); \(3^{2x+2} -28(3^x) + 3 = 0\)
14. Solve for x and y;
15. \(x\); \(\frac{9^y}{81} = \frac{9}{3^x} \)  and  \(\frac{27^x}{3^{y -8}} = \frac{1}{9} \)
16. Solve simultaneously; \(32^{x -1} × 8^y × \frac{1}{16} = 8\)  and  \(2^x × \frac{1}{8^y} = 64\)

Logarithms

This aspect of logarithm deals with applying the relationship between indices and logarithm and laws of logarithm to solve problems. The logarithm of any number to a given base is the index or power to which the base must be raised so that it is equal to that number. If \(N = b^x\), then \(log_b⁡N=x\), that is, the logarithm to base \(‘b’\) of a number \(N\), is the index (power \(x\)) to which ‘\(b\)’ must be raised to be equal to \(N\). The relations \(log_b⁡N=x\) and \(b^x=N\) are equivalent to each other. The form \(b^x=N\) is called the index form of writing numbers, while the form \(log_b⁡N=x\) is called the logarithmic form. This implies that for every logarithmic equation, there must be a corresponding index equation.

Examples:

1. \(log_{2⁡8}=3 ⇒2^3=8 \)

2. \(log_{10}⁡ 1000=3 ⇒ 10^3=1000\)

3. \(log_3⁡81=4 ⇒ 3^4=81\)

4. \(log_2⁡ \sqrt{8} = \frac{3}{2} ⇒ 2^{\frac{2}{3}} = \sqrt{8}\)

5. \(log_2⁡ \frac{1}{32} = -5 ⇒ 2^{-5}= \frac{1}{32}\)

The definition of \(log_a⁡x\) can also be stated as follows:

For any \(a > 0\)  and  \(a≠1\)

\(y = log_a⁡x\) if and only if \(a^y=x\)

NB: The base of a logarithm must be a positive number and cannot be \(1\).

Examples:

Evaluate the following without using tables or calculator.

(a) \(log_4⁡0.0625 \)

(b) \(log_{0.25}⁡128\)

(c) \(log_{\sqrt[2]{2}}⁡\sqrt{128}\)

(d) \(log_{10}⁡0.001\)

Solution:

(a) \(log_4⁡0.0625\)

Let \(log_4⁡0.0625=x \)

Then \(4^x=0.0625 \\ 4^x=\frac{625}{10000}=\frac{1}{16} \\ 4^x=\frac{1}{4}^2 \\ 4^x=4^{-2} \)

Equating powers

\(∴ x= -2\)

(b) \(log_{0.25}⁡128 \)

Let \(log_{0.25}⁡128=y \\ (0.25)^y=128  \\ \Big(\frac{1}{4}\Big)^y=128  \\\Big(\frac{1}{2^2}\Big)^y=2^7  \\ 2^{-2y}=2^7\)

Equating powers

\(-2y=7 \\ ∴ y = \frac{-7}{2}\)

(c) \(log_{\sqrt[2]{2}}⁡\sqrt{128}\)

Let \(log_{\sqrt[2]{2}}⁡\sqrt{128} = p \\ (\sqrt[2]{2})^p=⁡\sqrt{128} \\ \Big(2^{\frac{3}{2}}\Big)^p =128^{\frac{1}{2}} \\ 2^{\frac{3p}{2}}=2^{\frac{7}{2}}\)

Equating powers

\(\frac{3p}{2}=\frac{7}{2} \)

Cross-multiply

\(6p=14 \\ \frac{6p}{6}=\frac{14}{6} \)

∴ \(p=\frac{7}{3}\)   or   \(2 \frac{1}{3}\)

Let \(log_{10}⁡0.001 = m\)

\(10^m=0.001 \\ 10^m=\frac{1}{1000} \\ 10^m=\frac{1}{10^3}  \\ 10^m=10^{-3} \\ ∴ m= -3\)

CLASS ACTIVITY

1. Express the following in index form.

(a) \(log_3⁡9=2\)

(b) \(log_3⁡x=10\)

(c) \(log_c⁡t=4\)

(d) \(log_5⁡0.04=-2\)

Express the following in logarithm form.

(a) \(p^3=m \)

(b) \(64=16^{\frac{3}{2}} \)

(c) \(\Big(\frac{1}{3}\Big)^{-3}=27 \)

(d) \(0.001=10^{-3} \)

(e) \(2^x=4\sqrt{2}\)

PRACTICE EXERCISES

Evaluate the following (without using tables or calculator) with the use of the relationship between indices and logarithms.

1. \(log_{12}⁡144\)

2. \(log_4⁡1\)

3. \(log_{0.5} \Big(\frac{1}{64}\Big)\)

4. \(log_{\sqrt{2}⁡2}\)

5. \(log_{10}\sqrt{10}\)

Laws of Logarithm

The following are the different rules that may be applied when solving problems on logarithms.

1. Addition Law

\(log_b⁡MN=log_b⁡M+log_b⁡N\)

This implies that when two logarithms of the same base are multiplied, the result is the addition of the two logarithms to their common base.

Example:

\(log_{10}⁡ 72 = log_{10⁡}(10×7.2)=log_{10}⁡10+log_{10⁡}7.2\)

2. Subtraction Law

\(log_b⁡\Big(\frac{M}{N}\Big) = log_b⁡M -log_b⁡N\)

This shows that when logarithms are divided, the logarithm of the denominator is subtracted from the logarithm of the numerator in their common base.

Example:

\(log_{10}⁡0.3=log_{10}⁡\Big(\frac{3}{10}\Big)=log_{10}⁡3 -log_{10⁡}10\)

3. Logarithm of its Own Base

The logarithm of any number to its own base is equal to \(1\).

Example:

\(log_5⁡5=1\), \(log_{\sqrt{7}}⁡sqrt{7}=1\), \(log_{\frac{2}{3}}\Big(\frac{2}{3}\Big)=1\), etc.

4. Power Law

\(log_b⁡N^k= klog_b⁡N\)

When a logarithm is raised to a certain power, the power is used to multiply the logarithm itself.

Example 1:

\(log_{10}⁡10^3=3log_{10}⁡10  \\ =3×1=3 \)

Note: \( log_a⁡a=1\)

Example 2:

\(log_2⁡8^3=3 log_2⁡8 \\ =3log_2⁡2^3 \\ =3×3 log_2⁡2 \\ =3×3×1\)

\(=9\) since \(log_2⁡2=1\)

5. Power of the Base

\(log_{a^2 }⁡x = \frac{1}{2} log_ax\)

When it is the base that has a power, the reciprocal of this power is used to multiply the logarithm itself.

Example 1:

\(log_{2^3}16=\frac{1}{3} log_2⁡16 \\ = \frac{1}{3} log_2⁡2^4  \\= \frac{4}{3} log_2⁡2 = \frac{4}{3}×1=\frac{4}{3}\)

Example 2:

\(log_{\sqrt{9}}27=log_{9^{\frac{1}{2}}}27 \\ = 2 log_9 ⁡27 \\ = 2 log_(3^2 )⁡27 \\ = 2× \frac{1}{2} log_3⁡27 \\ = log_3⁡3^3 \\= 3 log_3⁡3=3×1=3\)

6. Logarithm of 1

\(log_b⁡1=0 \)

The logarithm of 1 to any base is equal to zero i.e \(log_3⁡ 1=0\), \(log_10 ⁡1=0\), \(log_7 ⁡1=0 \)

Example:

If \(log_{10}⁡1=x \)

Then, \(10^x=1 \)

But we know that any number raised to power zero is 1

So that, \(10^x=10^0\)

∴ \(x=0\)

Showing that \(log_{10}⁡ 1=0\)

7. Reciprocal Law

\(log_a⁡x=\frac{1}{log_x⁡a}\)

When the reciprocal of a logarithm is required, the base and the number interchange their positions.

Example:

\(log_4⁡8=\frac{1}{log_8⁡4}\)

8. Change of Base

\(log_a⁡x=\frac{log_n⁡x}{log_n⁡a} \)

Example:

\(log_3⁡8=\frac{log_{10}⁡8}{log_{10}⁡3}\),   \(log_{\sqrt{2}⁡}16=\frac{log_{10}⁡16}{log_{10}\sqrt{2}}\), etc.

This shows that when the base of a logarithm is changed, the initial base is used as a separate logarithm to divide the initial logarithm to the new base.

General Examples

Simplify \(\frac{log\sqrt{27}}{log 9}\)

Solution:

\(\frac{log 27^{\frac{1}{2}}}{log 9} = \frac{log 3^{\frac{3}{2}}}{log 3^2} \\ = \frac{\frac{3}{2} log 3}{2 log 3} \\ = \frac{3}{2} ÷ \frac{2}{1} \\ = \frac{3}{2} × \frac{1}{2} \\ = \frac{3}{4}\)

2. Simplify \(log_{10}⁡ \sqrt{20} + log_{10}⁡5 -log_{10}⁡\sqrt{5}\)

Solution:

\(log_{10}⁡ 2\sqrt{5} + log_{10}⁡5 -log_{10}⁡\sqrt{5} = log_{10} \Big(\frac{2⁡\sqrt{5} × 5}{⁡\sqrt{5}}\Big) \\ = log_{10}(2 × 5) \\ = log_{10}10 \\ = 1\)

3. Evaluate \(x\) if \(log_10⁡10^x=3x+2\)

Solution:

\(10^x=10^{3x+2}\)

Equating powers

\(x=3x+2⇒x= -1\)

Find \(x\), if \(log_4⁡ (x+3)(x -3) =2\)

Solution:

\(log_4⁡ (x^2 -9) = 2 \\ (x^2 -9)=4^2 \\ (x^2 -9)=16 \\x^2=16+9=25 \\ x^2=5^2\)

We have to equate the base since the powers are the same

\(⇒x=5\)

Find the value of \(x\) given that \(log_{10}⁡5+log_{10}⁡(x+2) -log_{10}⁡(x -1)=2\) (WAEC SSCE)

Solution:

Using the combined laws of logarithm (addition and subtraction laws), we have

\(log_{10} \Big(\frac{5(x + 2)}{x -1}\Big) = 2 \\ \frac{5(x + 2)}{x -1} = 10^2 \\ 5(x+2)=100(x -1) \\ 5x+10=100x -100 \\5x -100x= -100 -10 \\ -95x= -110 \\ x=\frac{-110}{-95} \\ ∴ x=1 \frac{3}{19}\)

Try this yourself: Simplify \(log_{10}⁡10 -2log_{10} \Big( \frac{1}{5}\Big) -log_{10}⁡2.5\) (The answer is \(2\))

Substitution in Logarithm

Given that \(log⁡3=0.4771\), \(log⁡2=0.3010\), find the values of the following without using logarithm table. (NECO 2002 Q5)

(i) \(log⁡\sqrt{6} \)

(ii) \(log⁡\sqrt[3]{0.3}\)

Solution:

\(log⁡\sqrt{6} = log⁡6^{\frac{1}{2}} \\ = \frac{1}{2} log⁡6 = \frac{1}{2} log⁡ (2×3)\)

Applying the addition law of logarithm, we have

= \(\frac{1}{2}(log⁡2+log⁡3)\) But we were given \(log⁡2=0.3010\) and \(log⁡ 3=0.4771\)

= \(\frac{1}{2}(0.3010+0.4771) \\ =\frac{0.7781}{2} = 0.38905 \\ log⁡ sqrt[3]{0.3} = log⁡ (0.3)^{\frac{1}{3}} \\ = \frac{1}{3} log⁡ (0.3) \\ = \frac{1}{3} log⁡\Big(\frac{3}{10}\Big)  \\ = \frac{1}{3} (log⁡3 -log⁡{10})  \\ = \frac{1}{3} (log⁡3 -1) = \frac{1}{3}(0.4771 -1) \\ = \frac{-0.5229}{3} \\ = -0.1743\)

If \(log⁡2 = 0.3010\) and \(log⁡3=0.4771\), calculate without using tables the value of \(log ⁡0.72\)

Solution:

Log⁡ \(0.72 = log⁡\Big(\frac{72}{100}\Big) \\ = log 72 -log⁡100  \\ = log⁡ (8×9) -log⁡ 10^2  \\ = log⁡ 8 + log⁡9 -log⁡ 10^2  \\ = log⁡ 2^3 + log⁡ 3^2 -log⁡ 10^2  \\ =3 log⁡2 + 2log⁡3 -2log⁡10  \\ =3(0.3010) + 2(0.4771) -2(1)  \\ =0.9030+0.9542 -2  \\ = -0.1428\)

8. Given that \(log_{10}⁡ 5=0.699\) and \(log_{10}⁡ 3=0.4771\), find \(log_{10}⁡45\) without using mathematical tables, hence solve \(x^{0.8265}=45\)

Solution:

\(log_{10}⁡ 45 = log_{10}⁡ (9×5)  \\ = log_{10}⁡9+log_{10}⁡5  \\ = log_{10}⁡ 3^2  + log_{10}⁡ 5  \\ = 2 log_{10}⁡3+log_{10}⁡5  \\ = 2(0.4771)+(0.699)  \\ = 1.653\)

For \(x^{0.8265}=45\), we take logarithm to base ten of both sides;

i.e. \(log_{10} x^{0.8265}= log_{10}⁡45 \\ 0.8265 log_{10}⁡x=log_{10}⁡45 \)

But \(log_{10}⁡ 45=1.653\) from above

\(0.8265 log_{10}⁡x=1.653 \\ log_{10}⁡x=\frac{1.653}{0.8265} \\ log_{10}⁡x=2 \\ x = 10^2 \\ ∴ x=100\)

PRACTICE EXERCISES

1. Given that \(log⁡2=0.3010\),  \(log⁡3=0.4771\) and \(log⁡7=0.8451\), evaluate the following:

(a) \(log⁡63\)  (b) \(log⁡84\)  (c) \(log⁡20\)  (d) \(log⁡140\)

2. Evaluate the following as logarithm of single numbers.

(a) \(2 -log⁡25\)  (b) \(2 log⁡4 -log⁡8\)  (c) \(log⁡3+log⁡15 -log⁡5\)  (d) \(3 log⁡5+log⁡2 -log⁡50\)  (e) \(log⁡5+log⁡8 -1\)

ASSIGNMENT

1. Given that \(log⁡x -log⁡ (2x -1)=1\), find \(x\) (WAEC)

2. Evaluate without any tables \(3 log⁡2+log⁡20 -log⁡1.6\) (WAEC)

3. Solve the equation: \(log_2⁡ (x^2 -2) = log_2⁡ (x -1)+1\) (WAEC)

4. Evaluate \(log_{10} \sqrt{35}+log_{10}⁡ \sqrt{2} -log_{10}⁡ \sqrt{7}\) (WAEC)

5. Evaluate \(log_{10}⁡50+log_{10}⁡64 -log_{10}⁡32\) (WAEC)

6. If \(log_{10}(2x+1) -log_10⁡(3x -2)=1\), find \(x\)

7. If \(log_{10}(3x -1) -log_{10}⁡2=3\), find \(x \)

8. Simplify \(\frac{log\sqrt{8}}{log8}\)

9. Without using mathematical tables, find \(x\) given that \(6 log⁡ (x+4)=log⁡64 \)

10. If \(log_a⁡x=p\) and \(log_a⁡ y=q\), show that \(xy=a^{p+q}\), deduce that \(log_a ⁡xy =log_a x+log_a⁡y\)

KEYWORDS

logarithms, log, index, power, log, root, square, etc.