Question

Refer to the attachment.

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Answer 1

$\textbf{Answer :}$

Now,

1 + log2 (base 7)

= 1 + (log2)/(log7)

= (log7 + log2)/(log7)

= (log14)/(log7)

= log14 (base 7)

∴ (1/49)^{1 + log2 (base 7}

= {(1/7)²}^{log14 (base 7)}

= (1/7)^{log14 (base 7)}²

= (1/7)^{log196 (base 7)}

= [7^{log196 (base 7)}]^(- 1)

= 196^(- 1)

= 1/196

Again,

log7 (base 1/5)

= (log7)/(log 1/5)

= (log7)/(log1 - log5)

= (log7)/(- log5)

= - (log7)/(log5)

= - log7 (base 5)

∴ 5^[- log7 (base 1/5)]

= 5^[- {- log7 (base 5)}]

= 5^{log7 (base 5)}

= 7

∴ (1/49)^{1 + log2 (base 7}
+ 5^[- log7 (base 1/5)]

= 1/196 + 7

RULES :

a^{logM (base a)} = M

loga (base b) = (loga)/(logb) (base e)

#$\textbf{MarkAsBrainliest}$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\large\boxed{\sf{\left(\dfrac{1}{49}\right)^{1+\log _7\left(2\right)}+5^{-\log \:_{\tfrac{1}{5}}\left(7\right)}}}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{7+\dfrac{1}{196}}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

→ First solve  ${\sf{\left(\dfrac{1}{49}\right)^{1+\log _7\left(2\right)}}$

$\mathrm{Apply\:exponent\:rule}:\quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}$

$\sf{\left(\dfrac{1}{49}\right)^{1+\log _7\left(2\right)}=\dfrac{1^{1+\log _7\left(2\right)}}{49^{1+\log _7\left(2\right)}}}$

$\mathrm{Apply\:rule}\:1^a=1$

$\sf{1^{\log _7\left(2\right)+1}=1}$

$=\dfrac{1}{49^{1+\log _7\left(2\right)}}$

$\sf{49^{1+\log _{7}(2)}=2^{2} \cdot 49}$

$\sf{=\dfrac{1}{49^{1+\log _7\left(2\right)}}=\dfrac{1}{2^2\cdot \:49}}$

$\sf{2^2\cdot \:49=196}$

$\sf{\dfrac{1}{2^2\cdot \:49}}=\boxed{\sf{{\dfrac{1}{196}}}}}$

$\large\boxed{\sf{\left(\dfrac{1}{49}\right)^{1+\log _7\left(2\right)}}=\dfrac{1}{196}}}$

_______________________________

→ Now solve $\sf{5^{-\log \:_{\tfrac{1}{5}}\left(7\right)}}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\dfrac{1}{a^b}$

$5^{-\log _{\tfrac{1}{5}}\left(7\right)}=\dfrac{1}{5^{\log _{\tfrac{1}{5}}\left(7\right)}}$

$\sf{5^{\log _{\tfrac{1}{7}}(7)}=\dfrac{1}{7}}$

$=\dfrac{1}{\dfrac{1}{7}}$

$=\dfrac{7}{1}$

$\sf{=7}$

$\large{\boxed{\sf{5^{-\log \:_{\tfrac{1}{5}}\left(7\right)}=7}}}$

_______________________________

$\large\boxed{\sf{\left(\dfrac{1}{49}\right)^{1+\log _7\left(2\right)}+5^{-\log \:_{\tfrac{1}{5}}\left(7\right)}=\underline{\underline{7+\dfrac{1}{196}}}}}$