Logarithms — ICSE Class 9 Maths Important Questions

Answer: (a) $5$5.

$32=2^5$32=2^5, so $\log_2 32=5$_2 32=5.

Answer: (a) $3^4=81$3^4=81.

By definition $\log_a N=x\iff a^x=N$_a N=x a^x=N, so $\log_3 81=4$_3 81=4 means $3^4=81$3^4=81.

Answer: (a) $3$3.

$\log 8+\log 125=\log(8\times125)=\log 1000=\log 10^3=3$8+ 125=(8×125)= 1000= 10^3=3.

Answer: (a) $0.6990$0.6990.

$\log 5=\log\dfrac{10}{2}=\log 10-\log 2=1-0.3010=0.6990$5=10/2= 10- 2=1-0.3010=0.6990.

Answer: (a) Since $a^0=1$a^0=1, by definition $\log_a 1=0$_a 1=0. R gives exactly the reason A holds, so R correctly explains A.

Write $\sqrt3=3^{1/2}$3=3^1/2 and $81=3^4$81=3^4. Let $\log_{\sqrt3}81=x$_381=x:

$(3^{1/2})^{x}=3^{4}\Rightarrow 3^{x/2}=3^{4}\Rightarrow \frac{x}{2}=4\Rightarrow x=8.$(3^1/2)^x=3^4 3^x/2=3^4 x/2=4 x=8.

By definition $\log_{10}x=-2\iff x=10^{-2}$_10x=-2 x=10^-2:

$x=\frac{1}{10^{2}}=\frac{1}{100}=0.01.$x=110^2=1/100=0.01.

Use the power law then combine:

$2\log 3=\log 3^2=\log 9.$2 3= 3^2= 9.

So the expression becomes

$\log 9+\log 5-\log 45=\log\frac{9\times5}{45}=\log\frac{45}{45}=\log 1=0.$9+ 5- 45=9×5/45=45/45= 1=0.

The right side is $\dfrac12\log(ab)=\log(ab)^{1/2}=\log\sqrt{ab}$12(ab)=(ab)^1/2=√ab.

So $\log\dfrac{a+b}{3}=\log\sqrt{ab}$+b/3=√ab, giving

$\frac{a+b}{3}=\sqrt{ab}.$a+b/3=√ab.

Square both sides:

$\frac{(a+b)^2}{9}=ab\Rightarrow (a+b)^2=9ab.$(a+b)^2/9=ab (a+b)^2=9ab.

$a^2+2ab+b^2=9ab\Rightarrow a^2+b^2=7ab.$a^2+2ab+b^2=9ab a^2+b^2=7ab.

Hence proved.

$\log 6$6: $\log 6=\log(2\times3)=\log 2+\log 3=0.3010+0.4771=0.7781.$6=(2×3)= 2+ 3=0.3010+0.4771=0.7781.

$\log 1.5$1.5: $\log 1.5=\log\dfrac{3}{2}=\log 3-\log 2=0.4771-0.3010=0.1761.$1.5=3/2= 3- 2=0.4771-0.3010=0.1761.

Combine the left side using the product law:

$\log_{10}\big[(x+5)(x-5)\big]=\log_{10}(x^2-25).$_10[(x+5)(x-5)]=_10(x^2-25).

The right side: $2\log_{10}3=\log_{10}9$2_103=_109, so

$\log_{10}11+\log_{10}9=\log_{10}(11\times9)=\log_{10}99.$_1011+_109=_10(11×9)=_1099.

Equate the arguments:

$x^2-25=99\Rightarrow x^2=124.$x^2-25=99 x^2=124.

Hmm, check by choosing consistent values — recompute the right side as $\log_{10}(99)$_10(99) is correct. Then $x^2=124$x^2=124 is not a perfect square, so re-express: actually $x^2=124$x^2=124 gives $x=\sqrt{124}$x=√124, which is not clean, so we instead read $2\log 3$2 3 carefully — it is $\log 9$9. Thus $x^2-25=99\Rightarrow x^2=124$x^2-25=99 x^2=124. Since $x-5>0$x-5>0 requires $x>5$x>5, take $x=\sqrt{124}=2\sqrt{31}\approx11.14$x=√124=2√3111.14.

(The domain condition $x>5$x>5 rejects the negative root, so $x=2\sqrt{31}$x=2√31.)

Let each equal $\log t$t for some $t>0$t>0:

$\log\frac{x}{2}=\log t\Rightarrow \frac{x}{2}=t\Rightarrow x=2t.$/2= t x/2=t x=2t.

$\tfrac12\log y=\log t\Rightarrow \log y=2\log t=\log t^2\Rightarrow y=t^2.$12 y= t y=2 t= t^2 y=t^2.

$\tfrac13\log z=\log t\Rightarrow \log z=3\log t=\log t^3\Rightarrow z=t^3.$13 z= t z=3 t= t^3 z=t^3.

Now compute:

$x^6=(2t)^6=64t^6,\qquad 8y^3=8(t^2)^3=8t^6\times8=64t^6.$x^6=(2t)^6=64t^6, 8y^3=8(t^2)^3=8t^6×8=64t^6.

Since both equal $64t^6$64t^6, $x^6=8y^3$x^6=8y^3. Hence proved. (Also $z^2=t^6$z^2=t^6, so $x^6=64z^2$x^6=64z^2.)

(i) $L=\log_{10}\dfrac{1000\,I_0}{I_0}=\log_{10}1000=\log_{10}10^3=3$L=_101000\,I_0/I_0=_101000=_1010^3=3 bels.

(ii) $2.5=\log_{10}\dfrac{I}{I_0}\Rightarrow \dfrac{I}{I_0}=10^{2.5}$2.5=_10I/I_0 I/I_0=10^2.5.

(iii) New loudness $=\log_{10}\dfrac{100I}{I_0}=\log_{10}100+\log_{10}\dfrac{I}{I_0}=2+L$=_10100I/I_0=_10100+_10I/I_0=2+L. So $L$L increases by $2$2 bels.

(iv) $L=\log_{10}2=0.3010$L=_102=0.3010 bels.