Writing all terms with base (3), the total exponent is (8-3+8-10=3). Therefore, the value is \(3^{3}\), so choose the option \(3^{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(3^{2}\). Writing all terms with base (3), the total exponent is (8-3+8-10=3). Therefore, the value is \(3^{3}\), so choose the option \(3^{3}\).
Step 3
Exam Tip
सभी पदों को आधार (3) में लिखने पर कुल घात (8-3+8-10=3) नहीं बल्कि (3) है। इसलिए सही मान \(3^{3}\) है और विकल्पों में \(3^{3}\) चुनना चाहिए।
Since (12^{4}=\(2^{2}\cdot3\)^{4}=2^{8}\cdot3^{4}), division leaves \(2^{3}\cdot3\). In exams, prime-factorize first.
Step 2
Why this answer is correct
The correct answer is A. \(2^{3}\cdot3\). Since (12^{4}=\(2^{2}\cdot3\)^{4}=2^{8}\cdot3^{4}), division leaves \(2^{3}\cdot3\). In exams, prime-factorize first.
Step 3
Exam Tip
(12^{4}=\(2^{2}\cdot3\)^{4}=2^{8}\cdot3^{4}), इसलिए भाग देने पर \(2^{3}\cdot3\) बचता है। परीक्षा में पहले अभाज्य गुणनखंड करें।
Here \(2^{-3}+2^{-4}=\frac{1}{8}+\frac{1}{16}=\frac{3}{16}\), and \(2^{-5}=\frac{1}{32}\). Therefore, the value is \(\frac{3}{16}\div\frac{1}{32}=6\).
Step 2
Why this answer is correct
The correct answer is A. (6). Here \(2^{-3}+2^{-4}=\frac{1}{8}+\frac{1}{16}=\frac{3}{16}\), and \(2^{-5}=\frac{1}{32}\). Therefore, the value is \(\frac{3}{16}\div\frac{1}{32}=6\).
Step 3
Exam Tip
\(2^{-3}+2^{-4}=\frac{1}{8}+\frac{1}{16}=\frac{3}{16}\), और \(2^{-5}=\frac{1}{32}\)। इसलिए मान \(\frac{3}{16}\div\frac{1}{32}=6\) है।
Here (32^{\frac{2}{5}}=\(2^{5}\)^{\frac{2}{5}}=2^{2}=4), and (4^{-\frac{3}{2}}=\(2^{2}\)^{-\frac{3}{2}}=2^{-3}=\frac{1}{8}). The product is \(\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). Here (32^{\frac{2}{5}}=\(2^{5}\)^{\frac{2}{5}}=2^{2}=4), and (4^{-\frac{3}{2}}=\(2^{2}\)^{-\frac{3}{2}}=2^{-3}=\frac{1}{8}). The product is \(\frac{1}{2}\).
Step 3
Exam Tip
(32^{\frac{2}{5}}=\(2^{5}\)^{\frac{2}{5}}=2^{2}=4), और (4^{-\frac{3}{2}}=\(2^{2}\)^{-\frac{3}{2}}=2^{-3}=\frac{1}{8})। गुणनफल \(\frac{1}{2}\) 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).
Step 2
Why this answer is correct
The correct answer is A. (17). 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).
Step 3
Exam Tip
\(2^{a}=2^{4}\) 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।
Here (x^{9}=\(x^{3}\)^{3}=8) and (x^{6}=\(x^{3}\)^{2}=4), so the sum is (12). In exams, express powers as multiples of the given power.
Step 2
Why this answer is correct
The correct answer is A. (12). Here (x^{9}=\(x^{3}\)^{3}=8) and (x^{6}=\(x^{3}\)^{2}=4), so the sum is (12). In exams, express powers as multiples of the given power.
Step 3
Exam Tip
(x^{9}=\(x^{3}\)^{3}=8) और (x^{6}=\(x^{3}\)^{2}=4), इसलिए योग (12) है। परीक्षा में दी हुई घात के गुणजों में अभिव्यक्ति लिखें।
Since \(9^{x-1}=3^{2x-2}\), the total exponent is (x+2x-2=3x-2). From \(243=3^{5}\), (3x-2=5), so \(x=\frac{7}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{7}{3}\). Since \(9^{x-1}=3^{2x-2}\), the total exponent is (x+2x-2=3x-2). From \(243=3^{5}\), (3x-2=5), so \(x=\frac{7}{3}\).
Step 3
Exam Tip
\(9^{x-1}=3^{2x-2}\), इसलिए कुल घात (x+2x-2=3x-2) है। \(243=3^{5}\) से (3x-2=5) और \(x=\frac{7}{3}\)।
Here \(49^{-1}=7^{-2}\) and \(343^{-2}=7^{-6}\), so \(\frac{7^{4}\cdot7^{-2}}{7^{-6}}=7^{8}\). In exams, dividing by a negative power adds the exponent.
Step 2
Why this answer is correct
The correct answer is A. \(7^{8}\). Here \(49^{-1}=7^{-2}\) and \(343^{-2}=7^{-6}\), so \(\frac{7^{4}\cdot7^{-2}}{7^{-6}}=7^{8}\). In exams, dividing by a negative power adds the exponent.
Step 3
Exam Tip
\(49^{-1}=7^{-2}\) और \(343^{-2}=7^{-6}\), इसलिए \(\frac{7^{4}\cdot7^{-2}}{7^{-6}}=7^{8}\)। परीक्षा में ऋणात्मक घात से भाग करने पर घात जुड़ती है।
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.
Step 2
Why this answer is correct
The correct answer is B. (2). 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.
Step 3
Exam Tip
\(5^{x+2}-5^{x}=25\cdot5^{x}-5^{x}=24\cdot5^{x}=600\), इसलिए \(5^{x}=25=5^{2}\)। परीक्षा में सामान्य घात को बाहर निकालें।
Writing all terms with base (2), the exponent is (7-6+12-8=5). In exams, first convert composite bases into prime bases.
Step 2
Why this answer is correct
The correct answer is B. \(2^{5}\). Writing all terms with base (2), the exponent is (7-6+12-8=5). In exams, first convert composite bases into prime bases.
Step 3
Exam Tip
सभी पदों को आधार (2) में लिखने पर घात (7-6+12-8=5) मिलती है। परीक्षा में संयुक्त आधारों को पहले अभाज्य आधार में बदलें।
The total exponent on the left is (x+x+2-3=2x-1), and \(32=2^{5}\), so (2x-1=5), giving (x=3). In exams, convert the whole expression into one power.
Step 2
Why this answer is correct
The correct answer is B. (3). The total exponent on the left is (x+x+2-3=2x-1), and \(32=2^{5}\), so (2x-1=5), giving (x=3). In exams, convert the whole expression into one power.
Step 3
Exam Tip
बाएँ पक्ष की कुल घात (x+x+2-3=2x-1) है, और \(32=2^{5}\), इसलिए (2x-1=5) से (x=3)। परीक्षा में पूरी अभिव्यक्ति को एक ही घात में बदलें।
Here (x^{6}=\(x^{2}\)^{3}=27) and (x^{4}=\(x^{2}\)^{2}=9), so the difference is (18). In exams, express powers using the given \(x^{2}\).
Step 2
Why this answer is correct
The correct answer is A. (18). Here (x^{6}=\(x^{2}\)^{3}=27) and (x^{4}=\(x^{2}\)^{2}=9), so the difference is (18). In exams, express powers using the given \(x^{2}\).
Step 3
Exam Tip
(x^{6}=\(x^{2}\)^{3}=27) और (x^{4}=\(x^{2}\)^{2}=9), इसलिए अंतर (18) है। परीक्षा में दी हुई घात \(x^{2}\) के रूप में लिखें।
Here (16^{-\frac{3}{4}}=\(2^{4}\)^{-\frac{3}{4}}=2^{-3}) and (8^{\frac{2}{3}}=\(2^{3}\)^{\frac{2}{3}}=2^{2}), so the value is \(2^{-1}=\frac{1}{2}\). In exams, multiply powers of powers.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). Here (16^{-\frac{3}{4}}=\(2^{4}\)^{-\frac{3}{4}}=2^{-3}) and (8^{\frac{2}{3}}=\(2^{3}\)^{\frac{2}{3}}=2^{2}), so the value is \(2^{-1}=\frac{1}{2}\). In exams, multiply powers of powers.
Step 3
Exam Tip
(16^{-\frac{3}{4}}=\(2^{4}\)^{-\frac{3}{4}}=2^{-3}) और (8^{\frac{2}{3}}=\(2^{3}\)^{\frac{2}{3}}=2^{2}), इसलिए मान \(2^{-1}=\frac{1}{2}\) है। परीक्षा में घात के ऊपर घात को गुणा करें।
From \(2^{a}=2^{3}\), (a=3), and from \(3^{b}=3^{4}\), (b=4), so \(a^{b}=3^{4}=81\). In exams, compare powers using equal bases.
Step 2
Why this answer is correct
The correct answer is C. (81). From \(2^{a}=2^{3}\), (a=3), and from \(3^{b}=3^{4}\), (b=4), so \(a^{b}=3^{4}=81\). In exams, compare powers using equal bases.
Step 3
Exam Tip
\(2^{a}=2^{3}\) से (a=3) और \(3^{b}=3^{4}\) से (b=4), इसलिए \(a^{b}=3^{4}=81\)। परीक्षा में घातों की तुलना समान आधार पर करें।
Since \(125=5^{3}\), (x=3), and since \(32=2^{5}\), (y=5), so (x+y=8). In exams, write numbers as powers of their prime bases.
Step 2
Why this answer is correct
The correct answer is C. (8). Since \(125=5^{3}\), (x=3), and since \(32=2^{5}\), (y=5), so (x+y=8). In exams, write numbers as powers of their prime bases.
Step 3
Exam Tip
\(125=5^{3}\) से (x=3) और \(32=2^{5}\) से (y=5), इसलिए (x+y=8)। परीक्षा में संख्याओं को उनके मूल आधार की घात में लिखें।
Here (\(2^{5}\)^{3}=2^{15}), \(4^{-2}=2^{-4}\), and \(8^{2}=2^{6}\), so the net exponent is (15-4-6=5). In exams, convert all bases to (2).
Step 2
Why this answer is correct
The correct answer is A. \(2^{5}\). Here (\(2^{5}\)^{3}=2^{15}), \(4^{-2}=2^{-4}\), and \(8^{2}=2^{6}\), so the net exponent is (15-4-6=5). In exams, convert all bases to (2).
Step 3
Exam Tip
(\(2^{5}\)^{3}=2^{15}), \(4^{-2}=2^{-4}\), और \(8^{2}=2^{6}\), इसलिए कुल घात (15-4-6=5) है। परीक्षा में सभी आधार (2) में बदलें।
Here (\left\(9^{2}\right\)^{3}=\(3^{2}\)^{6}=3^{12}), and \(3^{12}\div3^{10}=3^{2}\). In exams, write (9) as \(3^{2}\).
Step 2
Why this answer is correct
The correct answer is B. \(3^{2}\). Here (\left\(9^{2}\right\)^{3}=\(3^{2}\)^{6}=3^{12}), and \(3^{12}\div3^{10}=3^{2}\). In exams, write (9) as \(3^{2}\).
Step 3
Exam Tip
(\left\(9^{2}\right\)^{3}=\(3^{2}\)^{6}=3^{12}), और \(3^{12}\div3^{10}=3^{2}\)। परीक्षा में (9) को \(3^{2}\) लिखें।
Since (\left\(\frac{27}{8}\right\)^{\frac{1}{3}}=\frac{3}{2}), (\left\(\frac{27}{8}\right\)^{-\frac{2}{3}}=\left\(\frac{3}{2}\right\)^{-2}=\frac{4}{9}). In exams, take the cube root first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4}{9}\). Since (\left\(\frac{27}{8}\right\)^{\frac{1}{3}}=\frac{3}{2}), (\left\(\frac{27}{8}\right\)^{-\frac{2}{3}}=\left\(\frac{3}{2}\right\)^{-2}=\frac{4}{9}). In exams, take the cube root first.
Step 3
Exam Tip
(\left\(\frac{27}{8}\right\)^{\frac{1}{3}}=\frac{3}{2}), इसलिए (\left\(\frac{27}{8}\right\)^{-\frac{2}{3}}=\left\(\frac{3}{2}\right\)^{-2}=\frac{4}{9})। परीक्षा में पहले घनमूल निकालें।
Here \(2^{x+1}+2^{x}=2\cdot2^{x}+2^{x}=3\cdot2^{x}=48\), so \(2^{x}=16=2^{4}\). In exams, factor the common power \(2^{x}\).
Step 2
Why this answer is correct
The correct answer is B. (4). Here \(2^{x+1}+2^{x}=2\cdot2^{x}+2^{x}=3\cdot2^{x}=48\), so \(2^{x}=16=2^{4}\). In exams, factor the common power \(2^{x}\).
Step 3
Exam Tip
\(2^{x+1}+2^{x}=2\cdot2^{x}+2^{x}=3\cdot2^{x}=48\), इसलिए \(2^{x}=16=2^{4}\)। परीक्षा में सामान्य घात \(2^{x}\) बाहर लें।
We have (\left\(3^{x}\right\)^{2}=3^{2x}) and \(729=3^{6}\), so (2x=6) and (x=3). In exams, rewrite both sides with the same base.
Step 2
Why this answer is correct
The correct answer is B. (3). We have (\left\(3^{x}\right\)^{2}=3^{2x}) and \(729=3^{6}\), so (2x=6) and (x=3). In exams, rewrite both sides with the same base.
Step 3
Exam Tip
(\left\(3^{x}\right\)^{2}=3^{2x}) और \(729=3^{6}\), इसलिए (2x=6) और (x=3)। परीक्षा में दोनों पक्षों को समान आधार में लिखें।
After cancellation, the denominator becomes \(2^8\cdot 5^2\). The larger exponent is (8), so the decimal terminates after (8) places.
Step 3
Exam Tip
The numerator may cancel powers of (5), but a larger power of (2) may still remain. चरण 1: \(625=5^4\) है। चरण 2: कटौती के बाद हर \(2^8\cdot 5^2\) बचेगा। बड़ी घात (8) है, इसलिए दशमलव (8) स्थानों पर समाप्त होगा। चरण 3: अंश में (5) की घात कटेगी, पर (2) की बड़ी घात रह सकती है।
The denominator has only (2) and (5), so the decimal terminates.
Step 2
Why this answer is correct
The exponents are (3) and (2), and the larger one is (3).
Step 3
Exam Tip
Exam tip: Maximum decimal places equal the larger exponent. चरण 1: हर में केवल (2) और (5) हैं, इसलिए दशमलव समाप्त होगा। चरण 2: घातें (3) और (2) हैं, बड़ी घात (3) है। चरण 3: परीक्षा सुझाव: अधिकतम दशमलव स्थान बड़ी घात के बराबर होते हैं।
