Question

If 2, $lo{g}_{3^x-4}4,lo{g}_{3^x+\frac{7}{2}}4$ are in HP, then x is equal to

• A. 1
• B. 2
• C. 4
• D. 0

Answer 1

The correct option is B 2

To find the value of $x$

Given , $2,$ $lo{g}_{3^x-4}4,lo{g}_{3^x+\frac{7}{2}}4$ are in HP ,

If the three terms $a,b,c$ are in HP then ,

$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

Here ,

$2,lo{g}_{3^x-4}4,lo{g}_{3^x+\frac{7}{2}}4$ are in HP

$\frac{2}{lo{g}_{3^x-4}4}=\frac{1}{2}+\frac{1}{lo{g}_{3^x+\frac{7}{2}}4}$

Now, using Property of logarthmic
$\frac{1}{lo{g}_{a}b}=lo{g}_{b}a$
The above equation becomes
$2lo{g}_4(3^x-4)=\frac{1}{2}+lo{g}_4(3^x+\frac{7}{2})$

$2lo{g}_4(3^x-4)-lo{g}_4(3^x+\frac{7}{2})=\frac{1}{2}$
$lo{g}_4(3^x-4{)}^2-lo{g}_4(3^x+\frac{7}{2})=\frac{1}{2}$

Again using the Property of logarthmic , we have

$lo{g}_{c}a-lo{g}_{c}b=lo{g}_c\left(\frac{a}{b}\right)$
$lo{g}_4\left(\frac{(3^x-4{)}^2}{(3^x+\frac{7}{2})}\right)=\frac{1}{2}$

$\left(\frac{(3^x-4{)}^2}{(3^x+\frac{7}{2})}\right)=4^{1/2}$

$\left(\frac{(3^x-4{)}^2}{(3^x+\frac{7}{2})}\right)=2$

On simplifying this we get

$(3^x{)}^2-10(3^x)+9=0$

On solving this equation ,we get

$3^x=9,3^x=1$

From this , we get

$x=2$ and $x=0$

But we cannot take $x=0$

because it doesn't satisfy the original equation

Finally , we take $x=2$

Therefore , $x=2$

Hence, Option B is correct