Relation, Function and Graphs Exercise: 2.4 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise 2.4

1) Prove that:

a) log_a(xy³/z²) = log_ax + 3log_ay – 2log_az

Solution:

L.H.S. = log_a(xy³/z²)

= log_a(xy³) – log_az².     [log_a(x/y) = log_ax – log_ay]

= log_ax +log_ay³– log_az²  [log_a(xy)= log_ax + log_ay]

= log_ax + 3log_ay – 2log_az [log_ax^p = P. log_ax]

= log_ax + 3log_ay – 2log_az

= R.H.S. 

b) log_a2x + 3(log_ax–log_ay) = log_a(2x⁴/y³)  [log_ax–log_ay = log_a(x/y)]

Solution:

 L.H.S. = log_a2x + 3(log_ax–log_ay)

= log_a2x + 3log_ax – 3log_ay

= log_a2x + log_ax³– log_ay³

₌ log_a(2x.x³) – log_ay³ [log_ax + log_ay = log_a(x/y)]

= log_a(2x⁴/y³)  [log_ax–log_ay = log_a(x/y)]

= R.H.S. 

c) log_ax²– 2log_a$\sqrt {\rm{x}} {\rm{\: }}$= log_ax

Solution:

L.H.S. = log_ax²– 2log_a$\sqrt {\rm{x}} {\rm{\: }}$

= log_ax² – 2log_ax^1/2

= 2log_ax – 2.$\frac{1}{2}$log_ax [log_ax^p = P. log_ax]

= 2log_ax –log_ax

= log_ax

= R.H.S.

d) $\[{a^{{{\log }_a}x}}\]$=x

Solution:

Proof:

Let log_ax = y

So, x = a^y

Or a^y = x

So, $\[{a^{{{\log }_a}x}}\]$= x   [ y = log_ax]

e) log_aa^x=x

Solution:

L.H.S. = log_aa^x

= x. log_a a [log_ax^p = P.log_ax]

=x

=RHS

f) (log a)² – (log b)²= log(ab).log(a/b)

Solution:

L.H.S. = (log a)² – (log b)²

= (log a + logb)(loga – logb)

= log(ab).log(a/b)

= R.H.S. 

g) log(1+2+3) = log1 + log2 + log3

Solution:

L.H.S. = log(1+2+3)

= log6

= log(1.2.3)

= log(1.2) + log3            [log (ab) = log a + logb]

= log1 + log2 + log3

= R.H.S.

h) x^{log y – log z}. y^{log z – logx}.z^{logx – logy}=1

Solution:

Let x^{log y – log z}. y^{log z – logx}.z^{logx – logy} = t …(i)

To show: t = 1.

Now, taking log on both sides of (i), we have,

log (x^{log y – log z}. y^{log z – log x} . z^{log x – log y}) = log t

or, log x^{log y – log z} + log y^{log z – log x} + log z^{log x – log y}  = log t

or, (log y – log z)log x + (log z – log x)log y + (log x – log y) log z = log t.

or, log y log x – log z log x + log z log y – log x log y + log x log z – log y log z = log t

or, 0 = log t

so, log t = log 1     [log 1 = 0]

So, t = 1 

i) (yz)^{log y – log z}. (zx)^{log z – logx}. (xy)^{logx – logy} = 1

Solution:

Let (yz)^{log y – log z}. (zx)^{log z – logx}. (xy)^{logx – logy} = t..(i)

To show: t = 1

Now, taking log on both sides of (i) we have,

log{(yz)^{log y – log z}. (zx)^{log z – logx}. (xy)^{logx – logy}} = log t

or, log(yz)^{log y – log z}. log (zx)^{log z – logx}. log (xy)^{logx – logy} = log t

or, (log y – log z)log(yz) + (log z – log x)log(zx) + (log x – log y)log(xy) = log t

or, (log y – log z)(log y + log z) + (log z – log x)(log z + log x) + (log x – log y)(log x + log y) = log t

or, (log y)² – (log z)² + (log z)² – (log x)² + (log x)² – (log y)² = log t

or. 0 = log t

so, log t = log 1     [log 1 = 0]

So, t = 1. Proved.

j) log $\sqrt {{\rm{a}}\sqrt {{\rm{a}}\sqrt {{{\rm{a}}^2}} } } $= 1

Solution:

L.H.S. = log $\sqrt {{\rm{a}}\sqrt {{\rm{a}}\sqrt {{{\rm{a}}^2}} } } $

= log $\sqrt {{\rm{a}}.\sqrt {{\rm{a}}.{\rm{a}}} } $

= log$\sqrt {{\rm{a}}\sqrt {{{\rm{a}}^2}} } $

= log $\sqrt {{\rm{a}}.{\rm{a}}} $

= log $\sqrt {{{\rm{a}}^2}} $

= log a

= 1

= R.H.S. 

2. If a²+b²=2ab, show that \[\log \frac{{a + b}}{2} = \frac{{\log a + \log b}}{2}\].

Solution:

Here: a²+b²=2ab

⇒ a²+b²-2ab=0

⇒ a-b=0 ⇒a=b

LHS: \[\log \frac{{a + b}}{2} = \log \frac{{a + a}}{2} = \log a\]

RHS: \[\frac{{\log a + \log b}}{2} = \frac{{\log ab}}{2} = \log \frac{{{a^2}}}{2} = \log \sqrt {{a^2}}  = \log a\]

Thus, LHS = RHS

3) If f(x) = log $\frac{{\left( {1 - {\rm{x}}} \right)}}{{1 + {\rm{x}}}}$ (-1<x<1), show that f $\left( {\frac{{2{\rm{ab}}}}{{1 + {{\rm{a}}^2}{{\rm{b}}^2}}}} \right)$ = 2f(ab) where |ab|<1.

Solution:

Here, f(x) = log $\frac{{\left( {1 - {\rm{x}}} \right)}}{{1 + {\rm{x}}}}$

Now,

f $\left( {\frac{{2{\rm{ab}}}}{{1 + {{\rm{a}}^2}{{\rm{b}}^2}}}} \right)$

= log $\frac{{1 - \frac{{2{\rm{ab}}}}{{1 + {{\rm{a}}^2}{{\rm{b}}^2}}}}}{{1 + \frac{{2{\rm{ab}}}}{{1 + {{\rm{a}}^2}{{\rm{b}}^2}}}}}$

= log $\frac{{1 + {{\rm{a}}^2}{{\rm{b}}^2} - 2{\rm{ab}}}}{{1 + {{\rm{a}}^2}{{\rm{b}}^2} + 2{\rm{ab}}}}$

= log $\left( {\frac{{1 - {\rm{a}}{{\rm{b}}^2}}}{{1 + {\rm{a}}{{\rm{b}}^2}}}} \right)$

= 2log $\left( {\frac{{1 - {\rm{a}}{{\rm{b}}^2}}}{{1 + {\rm{a}}{{\rm{b}}^2}}}} \right)$

= 2 f(ab)                 [log $\frac{{\left( {1 - {\rm{x}}} \right)}}{{1 + {\rm{x}}}}$ = f(x)]

 

4) If x = log_2a, y = log_3a,2a and z = log_4a3a, prove that xyz+1=2xyz.

Proof:

Here, x = log_2a ⇒  a = (2a)^x ..(i)

y = log_3a ⇒   2a = (3a)^y ..(ii)

and z = log_4a ⇒  3a = (4a)^z ..(iii)

Now, a = (2a)^x     [from(i)]

Or, a = (3a)^yx        [from(ii)]

Or, a = (4a)^xyz       [from(iii)]

Or, 4a.a = (4a).(4a)^xyz [Multiplying by 4a]

Or, (2a)² = (4a)^{xyz + 1}

Or, (3a)^2y = (4a)^{xyz + 1}    [2a = (3a)^y]

Or, (4a)^2yz = (4a)^{xyz + 1} [3a = (4a)^z]

So, 2z = xyz + 1

 

5) If $\frac{{{\rm{logx}}}}{{{\rm{y}} - {\rm{z}}}}$=$\frac{{{\rm{logy}}}}{{{\rm{z}} - {\rm{x}}}}$=$\frac{{{\rm{logz}}}}{{{\rm{x}} - {\rm{y}}}}$, prove that x^x.y^y.z^z = 1

Proof:

Let, $\frac{{{\rm{logx}}}}{{{\rm{y}} - {\rm{z}}}}$ = $\frac{{{\rm{logy}}}}{{{\rm{z}} - {\rm{x}}}}$= $\frac{{{\rm{logz}}}}{{{\rm{x}} - {\rm{y}}}}$ = k

So, log x = k (y – z)

Or, x log x = kx(y – z)

Or, log x^x = k (xy – zx) …(i)[logx^p = P logx}

Again, log y = k(z – x)

Or, y log y = ky(z – x)

So, log z^z = k (zx – yz) …(ii)

Again, log z = k(x – y)

Or, z log z = kz(x – y)

So, log z^z = k (zx –yz) …(iii)

Adding (i), (ii) and (iii).

Or, log x^x + log y^y + log z^z = k (xy – zx + yz – xy + zx – yz)

Or, log(x^x.y^y.z^z) = k.0 = 0

Or, log(x^x.y^y.z^z) = log 1                [log1 = 0]

So, x^x.y^y.z^z = 1 Proved.