Answer: \(x=4\).

Use the change of base formula:

\(\log_x10=\dfrac{\lg10}{\lg x}=\dfrac{1}{\lg x}.\)

Therefore

\(\dfrac{2}{\log_x10}=2\lg x.\)

The equation becomes

\(2\lg x-\lg(x+4)=\lg2.\)

Since \(2\lg x=\lg(x^2)\),

\(\lg(x^2)-\lg(x+4)=\lg2.\)

Using \(\lg A-\lg B=\lg\left(\dfrac{A}{B}\right)\),

\(\lg\left(\dfrac{x^2}{x+4}\right)=\lg2.\)

So

\(\dfrac{x^2}{x+4}=2.\)

Hence

\(x^2=2x+8,\)

so

\(x^2-2x-8=0.\)

Factorise:

\((x-4)(x+2)=0.\)

The question states \(x\gt0\), so \(x=-2\) is rejected.

Therefore

\(x=4.\)

Answer: (a) \(\log_2\left(\frac{8a^4}{b}\right)\). (b) \(x=100\) or \(x=\frac{1}{100}\).

Apply the laws of logarithms carefully, paying attention to the base of each logarithm and any restrictions on the argument.

For part (a), write \(3=\log_2 8\), and use the logarithm laws:

\(3+4\log_2a-\log_2b=\log_28+\log_2a^4-\log_2b.\)

Therefore

\(3+4\log_2a-\log_2b=\log_2\left(\frac{8a^4}{b}\right).\)

For part (b), use change of base:

\(\log_x10=\frac{1}{\lg x}.\)

The equation \(\lg x=4\log_x10\) becomes

\(\lg x=\frac{4}{\lg x}.\)

So \((\lg x)^2=4\), and hence \(\lg x=2\) or \(\lg x=-2\).

Thus \(x=100\) or \(x=10^{-2}=\frac{1}{100}\).

This gives the required answer: (a) \(\log_2\left(\frac{8a^4}{b}\right)\). (b) \(x=100\) or \(x=\frac{1}{100}\).

Answer: \(p=3\).

Apply the laws of logarithms carefully, paying attention to the base of each logarithm and any restrictions on the argument.

Use \(\frac{1}{\log_pa}=\log_ap\). The equation becomes

\(\log_a(p+1)+\log_ap-\log_a(p+2)+\log_a5=\log_a12.\)

Combining logarithms,

\(\log_a\left(\frac{5p(p+1)}{p+2}\right)=\log_a12.\)

Thus

\(\frac{5p(p+1)}{p+2}=12.\)

So \(5p^2+5p=12p+24\), giving

\(5p^2-7p-24=0.\)

Factorising,

\((5p+8)(p-3)=0.\)

The value \(p=-\frac85\) is not valid for the logarithms in the original equation, so \(p=3\).

This gives the required answer: \(p=3\).

Answer: \(rs=6561\).

We start with the main method. Rewrite all logarithms using a common base, then combine them using the product and power laws.

Use change of base for \(\log_9s\):

\(\log_9s=\frac{\log_3s}{\log_39}=\frac{\log_3s}{2}.\)

Therefore

\(2\log_9s=\log_3s.\)

The equation becomes

\(\log_3r+\log_3s=8.\)

Using the logarithm law,

\(\log_3(rs)=8.\)

Hence

\(rs=3^8=6561.\)

This completes the solution and gives the required result.

Answer: \(a=-\frac32\), \(b=\frac83\), \(c=-2\).

Answer: \(a=-\frac32\), \(b=\frac83\), \(c=-2\).

Simplify the numerator first:

\(\sqrt{p q^{4/3}}\,r^{-3}=p^{1/2}q^{2/3}r^{-3}.\)

The denominator is

\((pq^{-1})^2r^{-1}=p^2q^{-2}r^{-1}.\)

When dividing powers, subtract the exponents:

\(\frac{p^{1/2}q^{2/3}r^{-3}}{p^2q^{-2}r^{-1}}=p^{1/2-2}q^{2/3-(-2)}r^{-3-(-1)}.\)

Therefore

\(p^aq^br^c=p^{-3/2}q^{8/3}r^{-2}.\)

Hence \(a=-\frac32\), \(b=\frac83\), and \(c=-2\).

Answer: (a)(i) \(x\gt \frac35\). (ii) \(x=\frac{13}{5}\). (b) \(y=\frac{\sqrt[4]{x}}{6}\).

Answer: (a)(i) \(x\gt \frac35\). (ii) \(x=\frac{13}{5}\). (b) \(y=\frac{\sqrt[4]{x}}{6}\).

(a)(i) The logarithm \(\lg(5x-3)\) exists when its argument is positive:

\(5x-3\gt 0\).

Hence

\(x\gt \frac35\).

(a)(ii) From

\(\lg(5x-3)=1\),

we have

\(5x-3=10\).

Therefore \(5x=13\), so

\(x=\frac{13}{5}\).

(b) Rewrite the constant 4 as a logarithm to base \(y\):

\(4=\log_y(y^4)\).

Also,

\(\frac12\log_y64=\log_y\sqrt{64}=\log_y8\).

So

\(\log_y x=\log_y(y^4)+\log_y8+\log_y162\).

Combine the logarithms:

\(\log_y x=\log_y(1296y^4)\).

Thus

\(x=1296y^4\).

Since \(1296=6^4\),

\(y^4=\frac{x}{6^4}\).

As \(y\gt 0\),

\(\displaystyle y=\frac{\sqrt[4]{x}}{6}\).

Answer: \(\lg\left(\frac{1000a^2}{b^4}\right)\); \(a=2\) or \(a=64\).

The key point is to rewrite the equation using logarithm laws or a suitable substitution.

(a) Since \(3=\lg1000\),

\(3+2\lg a-4\lg b=\lg1000+\lg a^2-\lg b^4.\)

Using the laws of logarithms,

(b) Let \(u=\log_4a\). Then \(\log_a4=\frac1u\). The equation becomes

\(\frac3u+2u=7.\)

Multiplying by \(u\),

\(2u^2-7u+3=0.\)

Factorising,

\((2u-1)(u-3)=0.\)

So

\(u=\frac12\quad\text{or}\quad u=3.\)

Therefore

\(\log_4a=\frac12\quad\text{or}\quad \log_4a=3.\)

Hence \(a=4^{1/2}=2\) or \(a=4^3=64\).

Answer: (a) \(x=-\frac17\). (b) \(b=a^2\) or \(b=a^{-1}\).

(a) Write all terms as powers of \(3\):

\(9^{5x}=(3^2)^{5x}=3^{10x},\)

\(27^{x-2}=(3^3)^{x-2}=3^{3x-6},\)

and

\(243=3^5.\)

So

\(\frac{3^{10x}}{3^{3x-6}}=3^5.\)

This gives

\(3^{10x-(3x-6)}=3^5,\)

so

\(3^{7x+6}=3^5.\)

Equating powers,

\(7x+6=5.\)

Hence

\(\boxed{x=-\frac17}.\)

(b) Let

\(t=\log_a b.\)

Then

\(\log_a\sqrt b=\frac12\log_a b=\frac12t,\)

and

\(\log_b a=\frac{1}{\log_a b}=\frac1t.\)

The equation becomes

\(\frac12t-\frac12=\frac1t.\)

Multiply by \(2t\):

\(t^2-t=2.\)

So

\(t^2-t-2=0.\)

Factorise:

\((t-2)(t+1)=0.\)

Thus \(t=2\) or \(t=-1\). Since \(t=\log_a b\),

\(\log_a b=2\quad\text{or}\quad \log_a b=-1.\)

Therefore

\(\boxed{b=a^2\text{ or }b=a^{-1}}.\)

Answer: \(\lg\left(\frac{100x^3}{y}\right)\).

Use the logarithm rules

\(a\lg x=\lg x^a,\qquad \lg A+\lg B=\lg(AB),\qquad \lg A-\lg B=\lg\left(\frac{A}{B}\right).\)

Also,

\(2=\lg100.\)

Therefore

\(3\lg x+2-\lg y =\lg x^3+\lg100-\lg y.\)

So the expression is

\(\lg\left(\frac{100x^3}{y}\right).\)

Answer: \(p+2\), \(7p-4\), \(x=1.26\).

Since \(16=4^2\),

\(\log_4(16x)=\log_4 16+\log_4 x=2+p.\)

Also \(256=4^4\), so

\(\log_4\left(\frac{x^7}{256}\right)=\log_4(x^7)-\log_4 256=7p-4.\)

Now use the equation:

\((p+2)-(7p-4)=5.\)

Hence

\(6-6p=5,\qquad p=\frac16.\)

Since \(p=\log_4x\),

\(x=4^{1/6}=1.2599\ldots.\)

Therefore

\(\boxed{x=1.26}.\)

Answer: \(x=\pm2\); \(k=-1\); integral \(=-\dfrac12\cos(2x+5)-x+C\).

Since \(\lg(x^2-3)=0\),

\(x^2-3=10^0=1.\)

Hence

\(x^2=4,\)

so

\(\boxed{x=\pm2}.\)

For part (b), use the logarithm laws:

\(\ln a^{\sin(2x+5)}=\sin(2x+5)\ln a\)

and

\(\ln\left(\frac1a\right)=\ln(a^{-1})=-\ln a.\)

Therefore

\(\frac{\ln a^{\sin(2x+5)}+\ln\left(\frac1a\right)}{\ln a} =\frac{\sin(2x+5)\ln a-\ln a}{\ln a}.\)

So the expression is

\(\sin(2x+5)-1.\)

Thus

\(\boxed{k=-1}.\)

Now integrate:

\(\int\left(\sin(2x+5)-1\right)\,dx =-\frac12\cos(2x+5)-x+C.\)

Hence

\(\boxed{-\frac12\cos(2x+5)-x+C}.\)

Answer: (a)(i) \(1+p+2q\), (ii) \(3p-q-1\), (iii) \(\frac1p+\frac1q\). (b) \(x=\log_3\frac43\approx0.262\).

(a)(i)

\(\log_a(axy^2)=\log_a a+\log_a x+\log_a y^2.\)

So

\(\log_a(axy^2)=1+p+2q.\)

(a)(ii)

\(\log_a\left(\frac{x^3}{ay}\right)=\log_a x^3-\log_a a-\log_a y.\)

Therefore

\(\log_a\left(\frac{x^3}{ay}\right)=3p-1-q.\)

(a)(iii) By the reciprocal law,

\(\log_x a=\frac1{\log_a x}=\frac1p,\qquad \log_y a=\frac1{\log_a y}=\frac1q.\)

Hence

\(\log_x a+\log_y a=\frac1p+\frac1q.\)

(b) Let

\(m=3^x.\)

Then

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

The equation becomes

\(m-3m^2+4=0.\)

So

\(3m^2-m-4=0.\)

Factorise:

\((3m-4)(m+1)=0.\)

Since \(m=3^x\gt 0\), we take

\(m=\frac43.\)

Therefore

\(3^x=\frac43,\)

so

\(\boxed{x=\log_3\frac43\approx0.262}.\)

Answer: (a) \(\log_3(pq)\). (b) \(a=5\) or \(a=\sqrt[3]{5}\).

(a) Use change of base:

\(\log_2p=\frac{\log_3p}{\log_32}.\)

So

\((\log_2p)(\log_32) = \left(\frac{\log_3p}{\log_32}\right)(\log_32) = \log_3p.\)

Therefore

\((\log_2p)(\log_32)+\log_3q = \log_3p+\log_3q.\)

Using \(\log A+\log B=\log(AB)\),

\(\boxed{\log_3(pq)}.\)

(b) Let

\(u=\log_a5.\)

The equation becomes

\(u^2-4u+3=0.\)

Factorise:

\((u-1)(u-3)=0.\)

So

\(\log_a5=1 \quad\text{or}\quad \log_a5=3.\)

If \(\log_a5=1\), then \(a=5\).

If \(\log_a5=3\), then

\(a^3=5,\)

so

\(a=\sqrt[3]{5}.\)

Answer: (a) \(\log_3(pq)\). (b) \(a=5\) or \(a=\sqrt[3]{5}\).

(a) Use change of base:

\(\log_2p=\frac{\log_3p}{\log_32}.\)

So

\((\log_2p)(\log_32) = \left(\frac{\log_3p}{\log_32}\right)(\log_32) = \log_3p.\)

Therefore

\((\log_2p)(\log_32)+\log_3q = \log_3p+\log_3q.\)

Using \(\log A+\log B=\log(AB)\),

\(\boxed{\log_3(pq)}.\)

(b) Let

\(u=\log_a5.\)

The equation becomes

\(u^2-4u+3=0.\)

Factorise:

\((u-1)(u-3)=0.\)

So

\(\log_a5=1 \quad\text{or}\quad \log_a5=3.\)

If \(\log_a5=1\), then \(a=5\).

If \(\log_a5=3\), then

\(a^3=5,\)

so

\(a=\sqrt[3]{5}.\)