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21 questions tagged with exponent_equation.
If (\left\(7^{x}\right\)^{2}\cdot7^{x-1}=16807), what is the value of (x)?
A. (2)
बाएँ पक्ष \(7^{2x}\cdot7^{x-1}=7^{3x-1}\) है और \(16807=7^{5}\)। इसलिए (3x-1=5) और (x=2)। / The left side is \(7^{2x}\cdot7^{x-1}=7^{3x-1}\), and \(16807=7^{5}\). Hence (3x-1=5), so (x=2).
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If \(16^{x}=1024\) and \(32^{y}=1024\), what is the value of (x+y)?
A. \(\frac{9}{2}\)
\(1024=2^{10}\), \(16^{x}=2^{4x}\) से \(x=\frac{5}{2}\), और \(32^{y}=2^{5y}\) से (y=2)। इसलिए योग \(\frac{9}{2}\) है। / Since \(1024=2^{10}\), \(16^{x}=2^{4x}\) gives \(x=\frac{5}{2}\), and \(32^{y}=2^{5y}\) gives (y=2). Hence the sum is \(\frac{9}{2}\).
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If \(3^{a}=81\) and \(9^{b}=729\), what is the value of \(a^{b}-b^{a}\)?
A. \(\frac{37}{8}\)
(a=4) और \(9^{b}=3^{2b}=3^{6}\) से (b=3) है। इसलिए \(a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17\), अतः विकल्पों में यह मान नहीं है। / We get (a=4), and \(9^{b}=3^{2b}=3^{6}\) gives (b=3). Thus \(a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17\), which is not among the options.
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If \(4^{x}+4^{x+1}+4^{x+2}=336\), what is the value of (x)?
B. (2)
सामान्य पद \(4^{x}\) लेने पर (4^{x}(1+4+16)=336) मिलता है। इसलिए \(21\cdot4^{x}=336\), \(4^{x}=16\), और (x=2)। / Factoring \(4^{x}\), we get (4^{x}(1+4+16)=336). Thus \(21\cdot4^{x}=336\), \(4^{x}=16\), and (x=2).
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If \(3^{x}\cdot27^{x-1}=243\), what is the value of (x)?
B. (2)
\(27^{x-1}=3^{3x-3}\), इसलिए कुल घात (x+3x-3=4x-3) है। \(243=3^{5}\), इसलिए (4x-3=5) और (x=2)। / Since \(27^{x-1}=3^{3x-3}\), the total exponent is (x+3x-3=4x-3). Since \(243=3^{5}\), (4x-3=5), so (x=2).
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If \(6^{x+1}-6^{x}=900\), what is the value of (x)?
B. (3)
\(6^{x+1}-6^{x}=6\cdot6^{x}-6^{x}=5\cdot6^{x}=900\), इसलिए \(6^{x}=180\) नहीं बनता। इसलिए दिए विकल्पों में कोई भी सही नहीं है। / Here \(6^{x+1}-6^{x}=5\cdot6^{x}=900\), so \(6^{x}=180\), which is not a listed integral power. Therefore none of the listed options is correct.
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If \(a\neq0\) and \(\frac{a^{3p-2}\cdot a^{p+5}}{a^{2p-1}}=a^{10}\), what is the value of (p)?
B. (3)
कुल घात ((3p-2)+(p+5)-(2p-1)=2p+4) है। (2p+4=10) से (p=3) मिलता है। / The total exponent is ((3p-2)+(p+5)-(2p-1)=2p+4). From (2p+4=10), we get (p=3).
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If (\left\(5^{x}\right\)^{2}\cdot5^{x-2}=3125), what is the value of (x)?
B. \(\frac{7}{3}\)
बाएँ पक्ष \(5^{2x}\cdot5^{x-2}=5^{3x-2}\) है और \(3125=5^{5}\)। इसलिए (3x-2=5) और \(x=\frac{7}{3}\)। / The left side is \(5^{2x}\cdot5^{x-2}=5^{3x-2}\), and \(3125=5^{5}\). Hence (3x-2=5), so \(x=\frac{7}{3}\).
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If \(9^{x}=729\) and \(27^{y}=729\), what is the value of (x+y)?
A. \(\frac{7}{2}\)
\(729=3^{6}\), \(9^{x}=3^{2x}\) से (x=3), और \(27^{y}=3^{3y}\) से (y=2)। इसलिए (x+y=5)। / Since \(729=3^{6}\), \(9^{x}=3^{2x}\) gives (x=3), and \(27^{y}=3^{3y}\) gives (y=2). Therefore, (x+y=5).
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If \(2^{a}=32\) and \(8^{b}=64\), what is the value of \(a^{b}-b^{a}\)?
B. (9)
(a=5) और \(8^{b}=2^{3b}=2^{6}\) से (b=2)। इसलिए \(a^{b}-b^{a}=25-32=-7\), अतः सही मान (-7) है। / We get (a=5), and from \(8^{b}=2^{3b}=2^{6}\), (b=2). Thus \(a^{b}-b^{a}=25-32=-7\), so the correct value is (-7).
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If \(3^{x}+3^{x+1}+3^{x+2}=117\), what is the value of (x)?
B. (2)
सामान्य पद \(3^{x}\) लेने पर (3^{x}(1+3+9)=117) मिलता है। इसलिए \(13\cdot3^{x}=117\), \(3^{x}=9\), और (x=2)। / Factoring \(3^{x}\), we get (3^{x}(1+3+9)=117). Thus \(13\cdot3^{x}=117\), \(3^{x}=9\), and (x=2).
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If \(2^{x}\cdot8^{x-2}=64\), what is the value of (x)?
B. (3)
\(8^{x-2}=2^{3x-6}\), इसलिए कुल घात (x+3x-6=4x-6) है। \(64=2^{6}\) से (4x-6=6) और (x=3)। / Since \(8^{x-2}=2^{3x-6}\), the total exponent is (x+3x-6=4x-6). From \(64=2^{6}\), (4x-6=6), so (x=3).
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If \(4^{x+1}-4^{x}=192\), what is the value of (x)?
B. (3)
\(4^{x+1}-4^{x}=4\cdot4^{x}-4^{x}=3\cdot4^{x}=192\), इसलिए \(4^{x}=64\)। \(64=4^{3}\), इसलिए (x=3)। / Here \(4^{x+1}-4^{x}=4\cdot4^{x}-4^{x}=3\cdot4^{x}=192\), so \(4^{x}=64\). Since \(64=4^{3}\), (x=3).
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If \(a\neq0\) and \(\frac{a^{2p+1}\cdot a^{p-3}}{a^{p+4}}=a^{6}\), what is the value of (p)?
C. (6)
कुल घात ((2p+1)+(p-3)-(p+4)=2p-6) है। (2p-6=6) से (p=6) मिलता है। / The total exponent is ((2p+1)+(p-3)-(p+4)=2p-6). From (2p-6=6), we get (p=6).
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If (\left\(3^{x}\right\)^{2}\cdot3^{x-1}=729), what is the value of (x)?
A. \(\frac{7}{3}\)
बाएँ पक्ष \(3^{2x}\cdot3^{x-1}=3^{3x-1}\) है और \(729=3^{6}\)। इसलिए (3x-1=6) और \(x=\frac{7}{3}\)। / The left side is \(3^{2x}\cdot3^{x-1}=3^{3x-1}\), and \(729=3^{6}\). Hence (3x-1=6) and \(x=\frac{7}{3}\).
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If \(6^{x}=216\) and \(36^{y}=216\), what is the value of (x+y)?
A. \(\frac{9}{2}\)
\(216=6^{3}\), इसलिए (x=3)। (36^{y}=\(6^{2}\)^{y}=6^{2y}=6^{3}), इसलिए \(y=\frac{3}{2}\) और योग \(\frac{9}{2}\) है। / Since \(216=6^{3}\), (x=3). Also (36^{y}=\(6^{2}\)^{y}=6^{2y}=6^{3}), so \(y=\frac{3}{2}\) and the sum is \(\frac{9}{2}\).
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If \(2^{a}=16\) and \(4^{b}=64\), what is the value of \(a^{b}-b^{a}\)?
A. (17)
\(2^{a}=2^{4}\) से (a=4), और \(4^{b}=4^{3}\) से (b=3)। इसलिए \(a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17\), अतः दिए विकल्पों में परिमाण (17) है। / From \(2^{a}=2^{4}\), (a=4), and from \(4^{b}=4^{3}\), (b=3). Thus \(a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17\), so the listed magnitude is (17).
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If \(2^{x}+2^{x+1}+2^{x+2}=112\), what is the value of (x)?
B. (4)
सामान्य पद \(2^{x}\) लेने पर (2^{x}(1+2+4)=112), इसलिए \(7\cdot2^{x}=112\)। इससे \(2^{x}=16\) और (x=4)। / Factoring \(2^{x}\), we get (2^{x}(1+2+4)=112), so \(7\cdot2^{x}=112\). Thus \(2^{x}=16\) and (x=4).
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If \(3^{x}\cdot9^{x-1}=243\), what is the value of (x)?
B. \(\frac{7}{3}\)
\(9^{x-1}=3^{2x-2}\), इसलिए कुल घात (x+2x-2=3x-2) है। \(243=3^{5}\) से (3x-2=5) और \(x=\frac{7}{3}\)। / Since \(9^{x-1}=3^{2x-2}\), the total exponent is (x+2x-2=3x-2). From \(243=3^{5}\), (3x-2=5), so \(x=\frac{7}{3}\).
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If \(5^{x+2}-5^{x}=600\), what is the value of (x)?
B. (2)
\(5^{x+2}-5^{x}=25\cdot5^{x}-5^{x}=24\cdot5^{x}=600\), इसलिए \(5^{x}=25=5^{2}\)। परीक्षा में सामान्य घात को बाहर निकालें। / Here \(5^{x+2}-5^{x}=25\cdot5^{x}-5^{x}=24\cdot5^{x}=600\), so \(5^{x}=25=5^{2}\). In exams, factor out the common power.
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If \(a\neq0\) and \(\frac{a^{p+4}\cdot a^{2p-1}}{a^{p+5}}=a^{8}\), what is the value of (p)?
C. (5)
कुल घात ((p+4)+(2p-1)-(p+5)=2p-2) है, इसलिए (2p-2=8) और (p=5)। परीक्षा में घातों को जोड़ते और घटाते समय चिह्न सावधानी से देखें। / The total exponent is ((p+4)+(2p-1)-(p+5)=2p-2), so (2p-2=8) and (p=5). In exams, watch signs while adding and subtracting exponents.
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