Solution To Exercises in Logarithm

The following exercises have been extracted from the New General Mathematics for Senior Secondary Schools Book 3. 
Solve the Following equations for $x$.
(a) $\log_{10}x=3$

We simply use the basic rule of logarithm to solve the question which is $P=\log_{b}N\Rightarrow{N}=b^P$

$=10^3=x$
$\Rightarrow{x}=1000$

(b)  $\log_{x}27=3$

$\Rightarrow{x}^3=27$

$\Rightarrow{x^3}=3^3$

$x=3$

(c)  $3^{x}=1$
 Remember from the laws of indices $a^0=1$
$3^x=3^0$
$x=0$

(d)  $4^{x+1}=8^x$

we simply equate the bases, so that we can cancel out the bases

$\Rightarrow{2^{2(x+1)}}=2^{3x}$
cancel out the bases
$2x+2=3x$
collect like terms and make $x$ subject of formula
$x=2$

(e)  $2^x\times{2^{3x-1}}=1$
by the law addition law of indices, this will become
$2^{x+(3x-1)}=1$

remeber $a^0=1$
$2^{4x-1}=2^0$
cancel out the bases
$4x-1=0$
$4x=1$
$x=\frac{1}{4}$

(f)   $2^{2x}-3\times{2^x}+2=0$
We collect like terms and use law of indices to simplify bases
$2^{2x+x}-3+2=0$
$2^{3x}-1=0$
$2^{3x}=1$
$2^{3x}=2^0$
$3x=0$
$x=0$

(g)  $6\log{(x+4)}=\log{64}$
Using the power law of Logarithm
$\log{(x+4)^6}=\log{2^6}$
$(x+4)^6=2^6$
$x+4=2$
$x=-2$

(h)  $9^{2x+1}=\frac{81^{x-2}}{3^x}$
We simply equate the bases of the L.H.S and R.H.S
$3^{2(2x+1)}=\frac{3^{4(x-2)}}{3^x}$
$3^{4x+2}=3^{3x-8}$
$4x+2=3x-8$
$x=-10$

(i)  $\log_{10}(2x+1)-\log_{10}(3x-2)=1$
Remember $\log{10^1}=1$
$\log_{10}(2x+1)-\log_{10}(3x-2)=\log{10}$
$\log_{10}(\frac{2x+1}{3x-2})=\log{10}$
$\frac{2x+1}{3x-2}=10$
Cross multiply
$2x+1=10(3x-2)$
$2x+1=30x-20$
$2x-30x+20+1=0$
$-28x+21=0$
$-28x=-21$
$x=\frac{21}{28}=\frac{3}{4}$

(j)  $3\times{9^{1+x}}=27^{-x}$
Simplify the bases in such a way that they are all the same
$3\times{3^{2(1+x)}}=3^{-3x}$
$3^{1+2+2x}=3^{-3x}$
Cancel out the bases
$1+2+2x=-3x$
Collect like terms
$3=-3x-2x$
$3=-5x$
$x=\frac{-3}{5}$

(k)  $\log_{10}(3x-1)-\log{10^2}=3$
Remember $\log{10^3}=\log{1000}=3$
$\log_{10}(\frac{3x-1}{2})=\log{10^3}$
$\frac{3x-1}{2}=1000$
Cross Multiply
$3x-1=2000$
Collect like terms
$3x=1999$
$x=\frac{1999}{3}$
$x=666.3$