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Exponents with Like Bases

Question

If ${(\frac{3}{2})}^{2} \times {(\frac{3}{2})}^{a + 5} = {(\frac{3}{2})}^{8}$ , then $a =$ ______ .
$- 1$
$1$
$2$
$0$

A

0

B

- 1

C

1

D

2

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Solution

Verified by Toppr

${(\frac{3}{2})}^{2} \times {(\frac{3}{2})}^{a + 5} = {(\frac{3}{2})}^{8} \Rightarrow {(\frac{3}{2})}^{a + 7} = {(\frac{3}{2})}^{8}$
On comparing, we get, $a + 7 = 8 \Rightarrow a = 1$

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Similar Questions

Q1

If

{(\frac{3}{2})}^{2} \times {(\frac{3}{2})}^{a + 5} = {(\frac{3}{2})}^{8}

, then

a =

______ .

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Q2

If

{(\frac{3}{2})}^{2} \times {(\frac{3}{2})}^{a + 5} = {(\frac{3}{2})}^{8}

, then the value of

a

is:

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Q3

If

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 2 & 3 & 4 0 & 5 & 6 \end{matrix} ⎤ ⎥ ⎦

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Q4

State whether True or False.
The H.C.F. and L.C.M. of:

2 a^{2} - a - 3, (2 a - 3)^{2}

and

4 a^{2} - 9

is

(2 a - 3); (2 a + 3) (a + 1) (2 a - 3)^{2}

If true then enter

1

and if false then enter

0

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Q5

If

A + 2 B = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 0 6 & - 3 & 3 - 5 & 3 & 1 \end{matrix} ⎤ ⎥ ⎦

and

2 A - B = ⎡ ⎢ ⎣ \begin{matrix} 2 & - 1 & 5 2 & - 1 & 6 0 & 1 & 2 \end{matrix} ⎤ ⎥ ⎦

, then

t r (A) - t r (B) =

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