LOGARITHM EQUATION 2 (7 KILLER QUESTION)

In my previous post, I introduce the logarithmic equation. Today, we would be solving killer logarithmic questions.

Ok, let's get started

Question 1
Evaluate $\log_{2}x^{16}+\log_{4}x^4=37-\log_{16}x^2$

Answer
Firstly, let's collect like the term
$\log_{2}x^{16}+\log_{4}x^4+\log_{16}x^2=37$

This translates to
$16\log_{2}x+4\log_{4}x+2\log_{16}x=37$

Recalled that $\log_{b}a$ can also be written as $\frac{\log_{c}b}{\log_{c}a}$
Therefore, $16\log_{2}x+4\log_{4}x+2\log_{16}x=37$
becomes
$16\log_{2}x+\frac{4\log_{2}x}{\log_{2}4}+\frac{2\log_{2}x}{\log_{2}16}=37$
$16\log_{2}x+\frac{4\log_{2}x}{\log_{2}2^2}+\frac{2\log_{2}x}{\log_{2}2^4}=37$
$16\log_{2}x+\frac{4\log_{2}x}{2\log_{2}2}+\frac{2\log_{2}x}{4\log_{2}2}=37$
$16\log_{2}x+\frac{4\log_{2}x}{2(1)}+\frac{\log_{2}x}{4(1)}=37$
$16\log_{2}x+\frac{4\log_{2}x}{2}+\frac{2\log_{2}x}{4}=37$

For simplicity, let's multiple through by $4$
$4(16\log_{2}x)+4(\frac{4\log_{2}x}{2})+4(\frac{(2\log_{2}x)}{4})=4\times 37$

This becomes
$64\log_{2}x+\frac{16\log_{2}x}{2}+\frac{8\log_{2}x}{4}=148$
$64\log_{2}x+8\log_{2}x+2\log_{2}x=148$
$74\log_{2}x=148$

Divide both sides by $74$
$\frac{\require{cancel}\bcancel{74}\log_{2}x}{\require{cancel}\bcancel {74}}=\frac{148}{74}$
$\log_{2}x= 2$

$x=2^2$
$x=4$

Logarithmic equations can also be in the form of simultaneous equations

Question 2
Solve the simultaneous logarithmic equation
$x=y^5$
$\log_{2}x-\log_{2}y=4$

Answer
$\log_{2}x-\log_{2}y=4$
Will become
$\log_{2}\frac{x}{y}=4$

Index form
$\frac{x}{y}=2^4$
$\frac{x}{y}=16$

Cross multiply
$x=y(16)$
$x=16y$

Substituting the value of $x$ in $x=y^5$
$16y=y^5$

Divide both sides by $y$
$\frac{16y}{y}=\frac{y^5}{y}$
$16=y^4$
$\sqrt[4]{16}=\sqrt[4]{y^4}$
$2=y$
$y=2$

$x$ can easily be gotten by substituting the value of y in $x=y^5$
$x=2^5$
$x=32$

Question 3
Solve for x in $\log_{4}x-\log_{x}16=1$

Answer
$\log_{4}x-\log_{x}4^2=1$
$\log_{4}x-2\log_{x}4=1$

Let $\log_{4}x$ be equal to $b$
$b-2\frac{(1)}{b}=1$
Note $\log_{4}x$ is the inverse of $\log_{x}4$
$b-\frac{2}{b}=1$
$\frac{b^2-2}{b}=1$

Cross multiply
$b^2-2=b$
$b^2-2-b=0$
$b^2-b-2=0$
$b^2-2b+b-2=0$
$b(b-2)+1(b-2)=0$
$(b+1)(b-2)=0$
$b=-1$ or $b=2$

Recalled,  $\log_{4}x=b$
When $b=-1$
$\log_{4}x=-1$

Index form
$x=4^{-1}$
$x=\frac{1}{4}$

when $b=2$
$\log_{4}x=2$
$x=4^2$
$x=16$ 
$x=16$ or $\frac{1}{4}$

Question 4
Evaluate $2\log_{2}y+\frac{1}{2}\log_{2}y=8-\log_{2}4$

Answer
$\log_{2}y^2+\log_{2}y^{\frac{1}{2}}=8+\log_{2}2^2$
$\log_{2}(y^2\times y^½)=8+2\log_{2}2$
$\log_{2}y^{2+\frac{1}{2}}=8+2(1)$
$\log_{2}y^{\frac{5}{2}}=8+2$
$\log_{2}y^{\frac{5}{2}}=10$

Index form
$y^{\frac{5}{2}}=2^{10}$
$y^{\frac{5}{2}}=1024$
$y^{\require{cancel}\bcancel{\frac{5}{2}}\times \require{cancel}\bcancel{\frac{2}{5}}}=1024^{\frac{2}{5}}$
$y=1024^{\frac{2}{5}}$
This is expressed as
$y=(\sqrt[5]{1024})^2$
$y=(4)^2$
$y=16$

Question 5
Solve the simultaneous equation
$\log_{2}x^2=\log_{2}y$
$\log_{2}(4x\sqrt{y})=6$

Answer
In a 1, we have the same base, so we will equate the log.
$x^2=y$....eqn 1

Now to Eqn 2
$\log_{2}(4x\sqrt{y})=6$

Index form
$4x\sqrt{y}=2^6$
$4x\sqrt{y}=64$

Substituting Eqn 1 ($y=x^2$) in Eqn 2
$4x\sqrt{x^2}=64$
Note, square root will cancels square
$4x(x)=64$
$4x^2=64$

Divide both sides by $4$
$\frac{\require{cancel}\bcancel{4}x^2}{\require{cancel}\bcancel{4}}=\frac{64}{4}$
$x^2=16$
$\sqrt{x^2}=\sqrt{16}$
$x=4$
Substituting the value of $x$ in Eqn 1
$(4)^2=y$
$16=y$
$y=16$

Related post

Question 6
Solve for x in the equation
$5\times5^{\log_{10}x}+5^{2-log_{10}x}=30$

Answer
Recalled $a^{2-3}=\frac{a^2}{a^3}$
Hence $5\times5^{\log_{10}x}+\frac{5^2}{5^{\log_{10}x}}=30$

Let $t=5^{\log_{10}x}$
$5t+\frac{5^2}{t}=30$
$\frac{5t^2+5^2}{t}=30$

Cross multiply
$5t^2+5^2=30t$

Dividing through by 5
$\frac{5t^2}{5}+\frac{5^2}{5}=\frac{30t}{5}$
$t^2+5=6t$
$t^2-6t+5=0$
$t^2-t-5t+5=0$
$t(t-1)-5(t-1)=0$
$(t-1)(t-5)=0$
$t=1$ or $t=5$

Recalled $t=5^{\log_{10}x}$
When $t=1$
$1=5^{\log_{10}x}$
$5^0=5^{\log_{10}x}$

Equating the power
$0=\log_{10}x$
$\log_{10}x=0$

Index form
$x=10^0$
$x=1$

When $t=5$
$5=5^{\log_{10}x}$
$5^1=5^{\log_{10}x}$
$1=\log_{10}x$
$\log_{10}x=1$
$x=10^1$
$x=10$
$x=10$, or $1$

Question 7
Given that $p^2+q^2=11pq$, where p and q are constant, show that $\frac{1}{2}(\log_{}p+\log_{}q)$ equals
a)$\log_{}(\frac{p-q}{3})$
b)$\log_{}(\frac{p+q}{\sqrt{13}}$)

Answer
Firstly, let's prove that $\frac{1}{2}(\log_{}p+\log_{}q)=\log_{}(\frac{p-q}{3})$

$p^2+q^2=11pq$

Subtract 2pq from both sides
$p^2+q^2-2pq=11pq-2pq$
$p^2+q^2-2pq=9pq$
$p^2-2pq+q^2=9pq$

By factorization
$p^2-1pq-1pq+q^2=9pq$
$p(p-q)-q(p-q)=9pq$
$(p-q)(p-q)=9pq$
$(p-q)^2=9pq$

Divide through by 9
$\frac{(p-q)^2}{9}=\frac{\require{cancel}\bcancel{9}pq}{\require{cancel}\bcancel{9}}$
$\frac{(p-q)^2}{9}=pq$

Taking the square root of both side
$\sqrt{\frac{(p-q)^2}{9}}=\sqrt{pq}$

This translate to
$\frac{\sqrt{(p-q)^2}}{\sqrt{9}}=\sqrt{pq}$
Square root cancels square, thus 
$\frac{p-q}{3}=\sqrt{pq}$
$\sqrt{pq}=\frac{p-q}{3}$

Now, let's go back to question (a)
$\frac{1}{2}(\log_{}p+\log_{}q)$
$\frac{1}{2}(\log_{}pq)$
This can be written as
$\log_{}\sqrt{pq}$
Recalled that
$\frac{p-q}{3}=\sqrt{pq}$
Hence $\log_{}\sqrt{pq}$ will be equal to
$\log_{}\frac{p-q}{3}$

Having proved (a), let's move to prove (b)
We want to prove that $\frac{1}{2}(\log_{}p+\log_{}q)=\log_{}(\frac{p+q}{\sqrt{13}}$

$p^2+q^2=11pq$
If $2pq-2pq=0$, then,
$p^2+q^2+2pq-2pq=11pq$
$p^2+q^2+2pq=11pq+2pq$
$p^2+q^2+2pq=13pq$
$p^2+2pq+q^2=13pq$

$p^2+pq+pq+q^2=13pq$
$p(p+q)+q(p+q)=13pq$
$(p+q)(p+q)=13pq$
$(p+q)^2$=13pq$

Divide through by 13
$\frac{(p+q)^2}{13}=\frac{\require{cancel}\bcancel{13}pq}{\require{cancel}\bcancel{13}}$
$\frac{(p+q)^2}{13}=pq$

Add square root to both sides
$\sqrt{\frac{(p+q)^2}{13}}=\sqrt{pq}$
$\frac{\sqrt{p+q)^2}}{\sqrt{13}}=\sqrt{pq}$
$\frac{p+q}{\sqrt{13}}=\sqrt{pq}$

Now to the log
$\frac{1}{2}(\log_{}p+\log_{}q)$ will give us $\log_{}\sqrt{pq}$
Let's insert the value of $\sqrt{pq}$
$\log_{}\frac{p+q}{\sqrt{13}}$

This is the end of our logarithm series. What have you learned? Is the post helpful? Tell me in the comment box. Alternatively, you can do so in our Facebook community.
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